For the following galvanic cell the emf will be at $25^{\large\circ}C$.$Ag/AgCl(s),KCl(0.2M\parallel KBr(0.001M).AgBr(s)/Ag$.The solubility product of $AgCl$ & $AgBr$ are $2.8\times 10^{-10}$ and $3.3\times 10^{-13}$ respectively.

Question

$\begin{array}{1 1}(a)\;-0.03V&(b)\;-0.1V\\(c)\;-0.049V&(d)\;-0.2V\end{array}$