suppose $A^{'}$ is sub module of $A$ and $$0\rightarrow A^{'}\overset{(d^{-1})^{'}}{\rightarrow}(I^{0})^{'} \overset{(d^{0})^{'}}{\rightarrow} (I^{1})^{'}\rightarrow \ldots $$
is injective resolution for $A^{'}$ .
1) show that we can make injective resolution $0\rightarrow A \overset{d^{-1}}{\rightarrow}I^{0} \overset{d^{0}}{\rightarrow}I^{1} \rightarrow ... $ for $A$ such that
i)$(I^{n})^{'} \subseteq I^{n} (n\geq0) $
ii)$coker((d^{n-1})^{'}) \subseteq coker (d^{n-1}) (n \geq 0)$
2) show that $$0 \rightarrow \frac{A}{A^{'}} \rightarrow \frac{I^{0}}{(I^{0})^{'}} \rightarrow \frac{I^{1}}{(I^{1})^{'}}\rightarrow \ldots $$
is injective resolution for $\frac{A}{A^{'}}$ .
this will be great if give me hint or Idea ,thanks a lot.
Hint: Try taking an injective resolution of $A/A^{\prime}$ and splicing the resolutions of $A^{\prime}$ and $A/A^{\prime}$ together to get an injective resolution of $A$. Note that condition (i) forces $(I^n)^{\prime}$ to be a summand of $I^n$, and together with (2) this already shows you how the splicing must look like termwise.