Yep. For every finite group, there's $M$ such that $H^2(G, M) = \Bbb Z/|G|$.
For every group (not necessary finite) augmentation ideal $I$ can be covered by free module of rank equal to rank of group: suppose $G$ generated by $g_i$, then map goes like $$\Bbb Z[G]^{\mathrm{rk} G} \to I, x_i \mapsto (g_i -1)$$
So we have short exact sequence $M \to \Bbb Z[G]^{\mathrm{rk} G} \to I$ where $M$ is kernel; by cohomological long exact sequence $$H^2(G, M) = H^1(G, I) = \Bbb Z/|G|$$
It's noteworthy that usually this module $M$ will have pretty big rank (we can bound it above by number of relations for some presentation of $G$ (see Lyndon, Schupp, Combinatorial group theory, Ch. II.3 on Fox calculus). It's an interesting question which conditions are implied on group by existense of cyclic (or $k$-generated) module which has $|G|$-torsion; I don't know answer to it.