 Solution

Choose $\hat{\zeta }\in H^2(\hat{\Gamma },M)$ with ${\iota }^{\ast }(\hat{\zeta })=\zeta$, and letbe its extension. By the given fact, $E$ is a profinite group, hence is residually finite. The extension $E_{\zeta }^{\prime }$ over $\Gamma$ is the pullback of $E$ along the injective map $\iota$. Part 5(a)(iii) embeds $E_{\zeta }^{\prime }$ into $E$. Since every subgroup of a residually finite group is residually finite, so is $E_{\zeta }^{\prime }$.