The equilibrium constant for the reaction Zn (s) + Sn ^ 2 + (aq) ⇌...

$Zn\left(s\right) + {Sn}^{2 +}\left(aq\right) ⇌ {Zn}^{2 +}\left(aq\right) + Sn\left(s\right)$

Given: $\dfrac{2 . 303RT}{F} = 0 . 059V$

${E}_{cell}^{0} = \dfrac{2 . 303RT}{2F}log{K}_{c}$

$⇒ {E}_{cell}^{0} = \dfrac{0 . 059}{2}\log ( {10}^{20} )$

$⇒ {E}_{cell}^{0} = \dfrac{0 . 059}{2} \times 20$

$⇒ {E}_{cell}^{0} = 0 . 59V$

${E}_{cell}^{0} = {E}_{{Sn}^{2 +} / Sn}^{0} - {E}_{{Zn}^{2 +} / Zn}^{0}$

$⇒ {E}_{cell}^{0} = {E}_{{Sn}^{2 +} / Sn}^{0} - ( - 0 . 76 )$

$⇒ {E}_{{Sn}^{2 +} / Sn}^{0} = - 0 . 76 + 059 = - 0 . 17$

$⇒ {E}_{Sn / {Sn}^{2 +}}^{0} = 0 . 17 = 17 \times {10}^{- 2}$

So, answer is 17.