We can see that the $\Delta G_1$,for the reaction (i) is zero as it is in equlibrium $\Delta G_2$ for the second reaction is $-nFE$ while for the third reaction is $-nFE^1$.Since $\Delta G$ is a thermodynamic quantity,it will follow the Fless's law,according to which all thermodynamic quantaties can be treated in the same manner as we treat the reaction.

That is to say if reaction $\Delta G_3=\Delta G_1+\Delta G_2$

$\Delta G_1$ is zero,$\Delta G_3=\Delta G_2$.This implies that $-nFE^1=-nFE$

$E^1=E$

This means that the half cell potential of third reaction is equal to that of second one.Now to calculate the half cell potential of $2^{nd}$ reaction.

$Ag^++e^-\rightarrow Ag$

We need its $E^0$ and concentration of $Ag^+$

$E^0$ is given and $[Ag^+]$ can be calculated as

$\large\frac{K_{sp}}{Cl^-}$

$\therefore E^1=E=E^0_{Ag^+/Ag}-0.059\log \large\frac{1}{[Ag^+]}$

$\Rightarrow E^0_{Ag^+/Ag}-0.059\log \large\frac{C_1}{K_{sp}}$

Hence (d) is the correct answer.