Let $k$ be a number field, $S$ a set of places of $k$, $k_S$ the maximal extension of $k$ unramified outside $S$, $\mathcal{O}_S$ the subring of $k_S$ with $\nu_{\mathfrak{p}}(\alpha)\geq 0$ for all $\mathfrak{p}\not\in S$, and $G_{S}$ the galois group $\text{Gal}(k_S/k)$. I am considering the following exercise at the end of chapter 8.3 in Cohomology of Number Fields.
Show the isomorphism $H^2(G_S,\mathcal{O}_S^\times)\cong Br(k_S/k)$
My attempt was to consider the following exact sequence
$$0\longrightarrow k_S^\times \longrightarrow I_{k_S}\longrightarrow C_{k_S}\longrightarrow 0$$
Where $I_{k_S}$ and $C_{k_S}$ are the Ideles and Idele class group of $k_S$. (Note that if $k_S$ is an infinite extension, these are simply the Galois invariance of the absolute Ideles and Idele class group.)
We take the cohomology long exact sequence of this exact sequence and obtain at $H^2$:
$$0\simeq H^1(G_S, C_{k_S})\longrightarrow H^2(G_S,k_S^\times)\simeq Br(k_S/k)\longrightarrow H^2(G_S,I_{k_S})\longrightarrow H^2(G_S, C_{k_S})$$
This gives an isomorphism $Br(k_S/k)\simeq \ker(H^2(G_S,I_{k_S})\to H^2(G_S, C_{k_S}))$. Now suppose $k=\mathbb{Q}$ and $S=\{3,\infty\}$, so $3^\infty|\#G_S$. Since $C$ is a formation module (defn. 3.1.8 in Cohomology of Number Fields) we have $H^2(G_S,C_{k_S})\simeq \frac{1}{\#G_S}\mathbb{Z}/\mathbb{Z}$. On the other hand, for $H^2(G_S,I_{k_S})$ we have
$$H^2(G_S,I_{k_S})(3)\simeq H^2(k,I)(3)\simeq \bigoplus_{\mathfrak p} H^2(k_{\mathfrak p})(3)\simeq \bigoplus_{\mathfrak p} \mathbb{Q}_3/\mathbb{Z}_3$$
by propositions $8.1.8$ and $8.1.7$ of the same book. (Note that we are using $A(p)$ to denote the $p$-power torsion of $A$.) Altogether, this yields
$$Br(k_S/k)(3)=H^2(G_S,k_S^\times)(3)\simeq \ker \left(\bigoplus_{\mathfrak p} \mathbb{Q}_3/\mathbb{Z}_3\longrightarrow \mathbb{Q}_3/\mathbb{Z}_3\right)$$
where the map is given by summation. Now by 8.3.11 of the same book, we have
$$H^2(G_S,\mathcal{O}_S^\times)(p)\simeq \ker\left(\bigoplus_{\mathfrak p\in S\setminus S_\infty} \mathbb{Q}_3/\mathbb{Z}_3\longrightarrow \mathbb{Q}_3/\mathbb{Z}_3\right)=\ker(\mathbb{Q}_3/\mathbb{Z}_3\to \mathbb{Q}_3/\mathbb{Z}_3)\simeq 0.$$
So these are two formulas, one for the $3$-part of $Br(k_S/k)$ and another for the $3$-part of $H^2(G_S,\mathcal{O}_S^\times)$. One is trivial and the other is decidedly not trivial, so they cannot be isomorphic.
Here is another approach I took, which similarly suggested that the groups are not isomorphic, or at least not canonically isomorphic. We have a canonical embedding $\mathcal{O}_S^\times\to k_S^\times$, and we would hope the isomorphism of their Brauer groups is induced by this embedding. To that end, suppose for simplicity that $k=\mathbb{Q}$ and $S=\{2,5,7,\dots,\infty\}$. we consider the following exact sequence
$$0\longrightarrow \mathcal{O}_S^\times \longrightarrow k_S^\times \longrightarrow \prod_{\mathfrak{p}|3} \mathfrak{p}^{\mathbb{Z}}=A$$
where the product is over the primes $\mathfrak{p}$ lying over $3$ and the last map need not be surjective. To turn this into a short exact sequence, we would like to determine the image of the last map.
To do this, fix a prime $\mathfrak{p}^*$ lying over $3$, and let $D\subseteq G_S$ be its decomposition group. The $A$ can be described as the set of functions $f:G_S\to \mathbb{Z}$ such that $f(\sigma\tau)=f(\tau)$ for all $\sigma \in D$. We identify each prime $\mathfrak{p}$ in the product with a coset of $G/H$, and then for $a\in A$ we set $f_a(H\sigma)=\nu_{(\mathfrak{p}^*)^\sigma}(a)$.
To determine the image of $k_S$ in $A$, note that anything in the image must be a continuous function. To see this, observe that any $\alpha \in k_S$ is in a Galois subextension $k_S/K/k$, and $f_\alpha$ then factors through the finite quotient $\text{Gal}(K/k)$ and is therefore continuous by properties of the Krull topology. Furthermore, all such continuous functions are possible: if $f\in A$ is continuous, then it factors through some finite quotient of $G_S$, and we can pass to the corresponding extension of $k$ to find a suitable $\alpha$.
With this description of the image of $k_S^\times$ in $A$ in mind, it becomes clear that we have constructed an isomorphism
$$k_S^\times/\mathcal{O}_S^\times\simeq \text{Ind}^D_{G_S}(\mathbb{Z})$$
since the collection of such continuous functions is precisely the definition of the induced module. Taking the long exact sequence in cohomology, we obtain
$$H^1(G_S,\text{Ind}^D_{G_S}(\mathbb{Z}))\longrightarrow H^2(G_S,\mathcal{O}_S^\times)\longrightarrow H^2(G_S,k_S^\times)\longrightarrow H^2(G_S,\text{Ind}^D_{G_S}(\mathbb{Z}))\longrightarrow H^3(G_S,\mathcal{O}_{S}^\times) $$
Computing the first and fourth term in this sequence, we obtain
$$H^1(G_S,\text{Ind}^D_{G_S}(\mathbb{Z}))\simeq H^1(D,\mathbb{Z})\simeq H^1(\hat{\mathbb{Z}},\mathbb{Z})\simeq \varinjlim H^1(\mathbb{Z}/n\mathbb{Z},\mathbb{Z})\simeq 0$$ $$H^2(G_S,\text{Ind}^D_{G_S}(\mathbb{Z}))\simeq H^2(D,\mathbb{Z})\simeq H^2(\hat{ \mathbb{Z}},\mathbb{Z})\simeq H^1(\hat{\mathbb{Z}},\mathbb{Q}/\mathbb{Z})\simeq \mathbb{Q}/\mathbb{Z}.$$
Taking the $p$-part of this exact sequence for $p\neq 3$ and using 8.3.11, we obtain
$$0\longrightarrow H^2(G_S,\mathcal{O}_S^\times)(p)\longrightarrow H^2(G_S,k_S^\times)(p)\longrightarrow \mathbb{Q}_p/\mathbb{Z}_p\longrightarrow 0.$$
Once again, we find that $H^2(G_S,\mathcal{O}_S^\times)$ is only a subobject of $H^2(G_S,k_S^\times)$, and similarly the failure for it to be an isomorphism comes from the primes not in $S$.
The ideas here apply more generally for other $S$ and $k$ and you still get a nontrivial cokernel of the inclusion in those cases.
I do not immediately see a flaw with either of these arguments, and the fact that they both show the exercise being false makes me think it might just be a mistake in the book, but I am far too aware of my shortcomings to believe this, so I am hoping someone can point out a mistake in these arguments.
