The cell reaction in this case are
At cathode :$Zn^{+2}+2e^-\rightarrow Zn(C_2)$
At anode :$Zn(C_1)\rightarrow Zn^{+2}+2e^-$
Net reaction : $Zn(C_1)\rightarrow Zn(C_2)$
$E=\large\frac{0.059}{2}$$\log \large\frac{C_2}{C_1}$
$\Rightarrow -0.0295\log (\large\frac{1}{2})$
$\Rightarrow 8.8\times 10^{-3}V$
Hence (d) is the correct answer.