I'm a bit confused about an exercise I read. Namely, T. Y. Lam's A First Course in Noncommutative Rings has the following on page $23$ Ex. $1.10$.
Let $p$ be a fixed prime. Show that there exists a noncommutative ring (with identity) of order $p^3$.
Well, this implies that there is a noncommutative ring with identity of order $8<16$, which contradicts with benh's answer given in Smallest non-commutative ring with unity . So have I understood correctly that there is a mistake in Lam's exercise?
Indeed, there is a noncommutative ring with unity of order $p^3$ for all prime $p$. Though I have not read benh's proof, it must have an error of some form. I will not bother to repeat the argument. See this paper for many various constructions related to your question as well as the proof that there is a noncommutative ring with unity of order $p^3$ for all prime $p$.
$(n-1) \times (n-1)$ matrices over ${\mathbb Z}_p$ of the form $a I + C e^T$ where $a \in {\mathbb Z}_p$, $C \in ({\mathbb Z}_p)^{n-1}$ (as a column vector) and $e^T = [1,0,\ldots,0]$ form a noncommutative ring (with identity) of order $p^n$. Thus for $n=4$ the matrices are of the form $$ \pmatrix{a_1 & 0 & 0\cr a_2 & a_3 & 0\cr a_4 & 0 & a_3\cr}$$
Here $p$ is an integer $\ge 2$ (not necessarily prime) and $n$ is an integer $\ge 3$. It is noncommutative because e.g. $$ \pmatrix{0 & 0 & \ldots\cr 0 & 1 &\ldots\cr \ldots & \ldots & \ldots} \pmatrix{0 & 0 & \ldots\cr 1 & 0 &\ldots\cr \ldots & \ldots & \ldots}= \pmatrix{0 & 0 & \ldots\cr 1 & 0 &\ldots\cr \ldots & \ldots & \ldots}$$ while $$ \pmatrix{0 & 0 & \ldots\cr 1 & 0 &\ldots\cr \ldots & \ldots & \ldots} \pmatrix{0 & 0 & \ldots\cr 0 & 1 &\ldots\cr \ldots & \ldots & \ldots}= \pmatrix{0 & 0 & \ldots\cr 0 & 0 &\ldots\cr \ldots & \ldots & \ldots}$$
