Error in calculating H+ activity
$$ \Delta G = \Delta G^\circ + RT \ln{\frac{a_{\ce{Sn^{4+}}}a_{\ce{H2O}}^{\;\;\;\;2}}{a_{\ce{SnO2}}a_{\ce{H+}}^{\;\;4}}} $$

At equilibrium, $\Delta G = 0$.

For $\Delta G^\circ = 2\times(-56.69) + 0.65 - (-123.20 - 4\times0) = +10.47\mathrm{\frac{kcal}{mol}}$ 

The question says "activity", but in the absence of any other information, let's assume that "activity" is equal to "concentration".  That is, let's assume activity coefficients are 1.

Thus, we get $$ \Delta G = 0 = +10.47  + RT \ln{\frac{a_{\ce{Sn^{4+}}}\cdot1^2}{1\cdot a_{\ce{H+}}^{\;\;4}}} $$

Thus, we get $$ {\left( \frac{-10.47}{RT}\right)}  = \ln{{10^{-10}} a_{\ce{H+}}^{\;\;-4}} = -4\ln{a_{\ce{H+}}} -10 \ln{10} $$

$$\frac{\frac{-10.47}{RT} + 10 \ln{10}}{-4} = \ln{a_{\ce{H+}}}$$