Total no of Faradays passed=$\large\frac{2\times 10^{-3}\times 16\times 60}{96500}$
$\Rightarrow 1.9\times 10^{-5}$
Moles of $Cu^{+2}$ ions deposited $=1.9\times \large\frac{10^{-5}}{2}$
Since absorbance was reduced to 50% of its original value the initial moles of $Cu^{+2}$ would be two times the moles of $Cu^{+2}$ reduced.
$\therefore$ Initial moles of $Cu^{+2}=\large\frac{1.9\times 10^{-5}}{2}$$\times 2$
$\Rightarrow 1.9\times 10^{-5}$
The concentration of $CuSO_4$ in solution =$1.9896\times 10^{-5}\times 4$
$\Rightarrow \approx 8\times 10^{-5}mol/lit$
Hence (b) is the correct answer.