I don't know if you still need the answer to the question, but I will answer it anyway. As you have probably realized by now, there is no easy answer to this question.
'How many possible $G$' is too difficult and I shall only be interested in 'How many possible extensions of $A$ by $N$'. Even with this simplification, it is a complicated problem.
There are two main ingredients to construct a group extension, you need a group automorphism $\phi: A\to Aut(N)$ and an element in $H^2(A,N_{\phi})$ which is the second cohomology group of $A$ with values in $N$ and the action given by $\phi$ (the definition of this object is not too difficult).
The answer to the question : 'How many possible extensions of $A$ by $N$ are there ?' is therefore given by
$$\sum_{\phi\in \hom(A,Aut(N))} \# H^2(A,N_{\phi})$$
and in all generalities, it is pretty much everything you can say.
If $A$ is cyclic, abstract considerations allow you to say that $H^2(A,N_{\phi})=H^0(A,N_{\phi})$ which is by definition the module $N^{\phi(A)}$ of elements in $N$ fixed by any automorphism in $\phi(A)$.
If $|A|$ and $|N|$ happen to be prime to each other, abstract considerations allow you to say $H^2(A,N_{\phi})=\{0\}$. As a result, in this case, you have as many extensions of $A$ by $N$ as $\# \hom(A,Aut(N))$. Furthermore for any $\phi\in \hom(A,Aut(N))$, the associated extension is a semi-direct product :
$$N\rtimes_{\phi}A\text{.} $$
If you are not in case $4$ or $5$, I doubt a lot that you can come to an understandable conclusion. If you are in case 5, I think you can even compute the isomorphism class.
For a reference about the link between group extensions and $2$-cocycles see the answer to this question.
Central extensions versus semidirect products
Here $H^2(A,N_{\phi})$ is some equivalence class of $2$-cocycles of $A$ with values in $N$, see the comment of Derek Holt to the aforementioned answer.