For the reaction of A + 2B-  ---- A2+ + 2B Kc is found to be 10^12. The E0 cell is:
1. 0.708 V
2. 0.1777 V
3. 0.088 V
4. 0.354 V

Dear Student, 
From the relation between free energy and electrode potential;
​​$∆G = -nF{E}_{cell}$
we know that at the equilibrium change in free energy is zero (ΔG) . hence E_cell becomes zero. 
From Nernst equation,
$E_{cell} = E^0 -\frac{0.0592}{n}\log Q$
Where Q is the reaction quotient.
As E_cell is zero we have, 
$E^0 = \frac{0.0592}{n}\log K_c$
Here we have replaced the Q by K_c as the reaction is in equilibrium. 
$E^{0 }= \frac{0.0592}{2}\log \left({10}^{12}\right) = 0.355 \approx 0.340 V$

Regards​