Here it is
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E_cell of the reaction under non standard conditions can be calculated by using Nernst equation.
E_cell = E⁰ cell -0.059/n *log Q
Q is reaction quotient.
H₂ +HB <----> HA + H₂

E⁰cell = E⁰ anode -E ⁰ cathode =0 V
  Here reduction and oxidation is of H₊ to H₂ and H₂ to H₊
n=2
E_cell = E⁰ cell -0.059 /2 log ( 1*10^-7)/(1*10^-5)

Ecell is the cell potential at non standard state conditions. Nernst equation is used to find Ecell using Eocell.

Ecell for given equation can be calculated as follows
Ecell = E0 cell - ( 0.059n) * log Q    ?  (Nernst equation)
Ecell = Eo - (0.0591/1) log ([HA]/[HB]
=0- 0.05911 log (10-710-5)
=0.1182V2
= 0.0591V or after rounding off we get 0.06

Hence option A is the correct one.