Since $20^9 = 2^{18} \cdot 5^9,$ requiring $a_1$ to divide $a_2$ and $a_2$ to divide $a_3$ says each prime’s exponents form a non-decreasing triple.

Non-decreasing triples are multisets: $\binom{21}{3}$ choices for the exponents of $2$ and $\binom{12}{3}$ for $5.$

Write $20^9 = 2^{18} \cdot 5^9,$ so each $a_i = 2^{x_i} 5^{y_i}$ with $0 \le x_i \le 18$ and $0 \le y_i \le 9;$ there are $19 \cdot 10 = 190$ divisors, and the exponents of the two primes are chosen independently and uniformly. The conditions that $a_1$ divides $a_2$ and $a_2$ divides $a_3$ hold exactly when $x_1 \le x_2 \le x_3$ and $y_1 \le y_2 \le y_3.$

Non-decreasing triples from a set of $k$ values correspond to multisets of size $3,$ counted by $\binom{k+2}{3}.$ So the probability is $$\begin{aligned} \frac{\binom{21}{3}}{19^3} \cdot \frac{\binom{12}{3}}{10^3} &= \frac{1330}{6859} \cdot \frac{220}{1000} \\ &= \frac{70}{361} \cdot \frac{11}{50} \\ &= \frac{77}{1805}. \end{aligned}$$

Since $1805 = 5 \cdot 19^2$ shares no factor with $77 = 7 \cdot 11,$ the fraction is in lowest terms and $m = 77.$