Counting

Firstly, let's prime factorize ${20}^{20}$:

${20}^{20}={(4\times5)}^{20}={2}^{40}\times{5}^{20}$

When we choose a factor of ${20}^{20}$, we are choosing a number of the form ${2}^{a}\times{5}^{b}$, where $0 \le a \le 40$ and $0 \le b \le 20$.

Since all divisors are equally likely to be chosen, a and b are randomized within their ranges.

If our factor is a multiple of 5, then we have $b \ge 1$. Otherwise, $b=0$, and our factor is not a multiple of 5.

Therefore, the probability that our factor is a multiple of 5 is the probability that $b \ge 1$.

There are 21 numbers between 0 and 20 inclusive, and there are 20 numbers from 1 to 20 inclusive, our answer is $\frac{20}{21}$.