jmc

The sum of the digits of $\frac{1}{5^{{}^n}}$ that are to the right of the decimal point is the sum of the digits of the integer $\frac{10^n}{5^{{}^n}}=2^n$, since multiplying by $10^n$ simply shifts all the digits $n$ places to the left. As a result, we start computing powers of 2, looking for an integer which has digits summing to a number greater than 10. $$\begin{array}{rl}{2}^1& =2\\ 2^2& =4\\ 2^3& =8\\ 2^4& =16\\ 2^5& =32\\ 2^6& =64\\ 2^7& =128\end{array}$$The sum of the digits in 128 is 11. The smallest positive integer $n$ such that the sum of the digits of $\frac{1}{5^{{}^n}}$ that are to the right of the decimal point is greater than 10 is $n=\boxed{7}$.