Consider the following hypothetical redox reaction

$ \text{A}(s)+2\text{B}^{+}(aq)\rightarrow\text{A}^{2+}(aq)+2\text{B}(s):E_{\text{cell}}^0=0.295\ \text{V} $

The equilibrium constant of the reaction at 298 K will be:

$ 10^{10} $

The correct answer is Option (2) → $ 10^{10} $

For the reaction 

$\text{A}(s)+2\text{B}^{+}(aq)\rightarrow\text{A}^{2+}(aq)+2\text{B}(s)\ E_{\text{cell}}^0=0.295$

This reaction shows that solid A is oxidized to $A^{2+}$ and $B^+$ ions are reduced to solid B

Total electrons transferred $n = 2$

The relationship between standard cell potential and equilibrium constant is:

$\Delta G^{\circ}=-nFE_{\text{cell}}^{\circ}$

$\Delta G^{\circ}=-RT\ln K$

Equating both:

$-nFE_{\text{cell}}^{\circ}=-RT\ln K\Rightarrow\ln K=\frac{nFE_{\text{cell}}^{\circ}}{RT}$

Substituting Values:

$n = 2$

$F=96500\ \text{C}/\text{mol}$

$E_{\text{cell}}^{\circ}=0.295$

$R=8.314\ \text{J}/\text{mol}\cdot\text{K}$

$T=298\ \text{K}$

$\ln K=(2\times96500\times0.295)/(8.314\times298)=22.98$

$K=10^{10}$