For each finite group $G$ there is a $G$-module $M$ that is a free abelian group of finite rank such that $H^2(G,M)=\mathbb{Z}/|G|$.

*Proof:* Let $I$ be the augmentation ideal of $\mathbb{Z}G$. Then $H^1(G,I)=\mathbb{Z}/|G|$. $\,\,I$ is a finitely generated $\mathbb{Z}G$-module (it's f.g. even as free abelian group). Hence there is a short exact sequence of $\mathbb{Z}G$-modules
$$0 \to M \to F \to I \to 0$$
where $F$ is a free $\mathbb{Z}G$-module of finite rank. In particular, $F$ is a free abelian group of finite rank and so is the subgroup $M$. The long exact cohomology sequence now yields $H^2(G,M) \cong H^1(G,I)$. q.e.d.

**Added:** If $\Omega^n(\mathbb{Z})$ denotes the kernel of $P_{n-1} \to P_{n-2}$ of a projective resolution $P \to \mathbb{Z}$ then the same argument shows $H^n(G,\Omega^n(\mathbb{Z}))=\mathbb{Z}/|G|$.