Solution:

The sum of the base-$10$ logs of the divisors is the log of their product. A number $N$ with $d(N)$ divisors has divisor product $N^{d(N)/2}.$

Here $N = 10^n$ has $d(N) = (n + 1)^2$ divisors, so the product is $(10^n)^{(n+1)^2/2}$ and its log is $$\frac{n(n + 1)^2}{2} = 792.$$

Thus $n(n + 1)^2 = 1584 = 11 \cdot 144 = 11 \cdot 12^2,$ giving $n = 11.$

Thus, the correct answer is A.