I claim that if $R$ is a left Noetherian ring and $M$ a finitely generated left $R$-module, then $M$ is injective as an $R$-module iff it is injective in the category of finitely generated $R$-modules.

Suppose $M$ is injective in the category of f.g. $R$-modules. Then for any left ideal $I$ of $R$ and any left $R$-module homomorphism $g: I \rightarrow M$, then since both $I$ and $R$ are finite $R$-modules and $M$ is injective over that category, $g$ extends to a map from $R$ to $M$. But then by Baer's criterion, $M$ is injective in the category of $R$-modules. The converse is trivial.

So when $R=\mathbb{Z}[G]$, since $R$ is not injective as an $R$-module, it is also not injective in the category of f.g. $R$-modules.

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I suspect the correct generalization should be "$R=\mathbb{Z}[G]$ has *finite injective dimension* as an $R$-module''. In other words, $R$ is a *Gorenstein* ring (assuming $R$ is Noetherian, which in this case it will be).

I don't know about the nonabelian case. However, if $G$ is a finite *abelian* group, then in Bass's seminal article (H. Bass, "On the ubiquity of Gorenstein rings" Math. Z. , 82 (1963) pp. 8–28), he shows that $R=\mathbb{Z}[G]$ is a *Gorenstein* ring of Krull dimension 1. In particular, he shows that $R$ has injective dimension $1$ as an $R$-module.

You can't do much better than this (again, in the commutative case). There's a classical result (cf. the book *Cohen-Macaulay rings*, by Bruns and Herzog, chapter 3) that says that if $R$ is a commutative Noetherian local ring and $M$ is a finitely generated $R$-module of finite injective dimension, then the injective dimension of $M$ equals the *depth* of $R$. (!) However, another classical result (same source) says that if $R$ is Gorenstein, then it must be Cohen-Macaulay -- that is, the depth of $R$ equals the Krull dimension of $R$. Put together, then, we have that if $R$ is Gorenstein, the injective dimension of $R$ over itself equals the Krull dimension of $R$.

Now, the rings you are looking at are not local; however, everything reduces to the local case. So again, as Jason was saying, the fact that $\mathbb{Z}$ has positive Krull dimension is what gets in the way of $\mathbb{Z}[G]$ being injective. It does, however, have injective dimension 1.