jmc

$$\begin{array}{rl}10^2-93& =7\\ 10^3-93& =907\\ 10^4-93& =9907\end{array}$$ This can be generalized into $10^n-93$ is equal is $n-2$ nines followed by the digits $07$. Then $10^{93}-93$ is equal to $91$ nines followed by $07$. The sum of the digits is equal to $9(91)+7=819+7=\boxed{826}$.