Let $G$ be a group and let $N$ be normal in $G$. Consider the ring homomorphism of group rings $\mathbb{Z}G\to\mathbb{Z}(G/N)$ defined by $g \mapsto gN$. Denote its kernel by $I_N'$. Let $I$ be the ideal $I_N(\mathbb{Z}G) = (\mathbb{Z}G)I_N$ where $I_N$ is the kernel of the augmentation map $$ \mathbb{Z}N \to \mathbb{Z}\;. $$ Clearly $I \leq I_N'$. Consider the $G$-module $\mathbb{Z}G/I$. If $x\in N$, since $(r+I)gx = rg+I$ (because $x-1$ is in $I$ as generator), then one can see $\mathbb{Z}G/I$ as a $G/N$-right module with the right multiplication $(r+I)gN=rg+I$.
Please explain me why $I_N'$ must act as $0$ for $\mathbb{Z}G/I$ (and so $I \geq I_N'$ too).
Note that a $G/N$-module $M$ is essentially nothing but a $G$-module on which $N$ acts trivially, that is, the ring homomorphism $\mathbb{Z}G \rightarrow \text{End}_{\mathbb{Z}}(M)^{\text{op}}$ defining the structure of right $\mathbb{Z}G$-module on $M$ factors as $\mathbb{Z}G \rightarrow \mathbb{Z}(G/N) \rightarrow \text{End}_{\mathbb{Z}}(M)^{\text{op}}$ and thus contains $I_N'$ in its kernel. As you already proved that $N$ acts trivially on $\mathbb{Z}G/I$, there is nothing left to do.