The equilibrium constant (K) for the reaction 

$\mathrm{Cu}\left(\mathrm{s}\right)+2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)⟶{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2\mathrm{Ag}\left(\mathrm{s}\right)$ will be 

We have 

                $E_{\text{cell }}^{\circ }=\frac{0.059}{n}\log K_c$

In the given reactions 

        $\begin{array}{r}\mathrm{Cu}\left(\mathrm{s}\right)⟹{\mathrm{Cu}}^{2+}+2e^{-}\\ 2{\mathrm{Ag}}^{+}+2e^{-}\rightleftharpoons 2\mathrm{Ag}\end{array}$

there is two electron change. 

So,     $n=2$

Substituting the values, in $\text{ (1) }$  we get:

      $0.46=\frac{0.0591}{2}\log K_c$

or     $\log K_c=\frac{2\times 0.46}{0.059}=15.6$

or     $K_c=\text{ Antilog 15.6 }$

So, the correct answer is $\left(d\right)$