Solution:

Using Kepler's 3rd Law $T^2 \propto r^3$, the ratio of time periods is $\frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2}$. Substituting the distances, we get $\frac{T_1}{T_2} = \left(\frac{10^{13}}{10^{12}}\right)^{3/2} = (10)^{3/2} = 10\sqrt{10}$.