Consider, for a positive integer $k$, the multiple $M_k = k \times 20$. If $k = 2^{a_1} \cdot 3^{a_2} \cdot 5^{a_3} \dots$ is the prime factorization of $k$, where $a_i \ge 0$ for all $i \ge 1$, then $M_k = 2^{a_1+2} \cdot 3^{a_2} \cdot 5^{a_3+1} \cdot 7^{a_4} \dots$ is the prime factorization of $M_k$.
For $M_k$ to have $20$ positive divisors, we need to have $(a_1+3)(a_2+1)(a_3+2) \times \prod_{i \ge 4} (a_i+1) = 20 = 2^2 \cdot 5$. This forces $a_1+3 \in \{4, 5, 10\}$ and $a_3+2 \in \{2, 4, 5\}$. Also, since we want the smallest such $k$, we may assume that $k$ has no prime divisors larger than $5$. (Any $a_i + 1 = 2$, with $i \ge 4$, can be replaced by $a_2 + 1 = 2$, resulting in a smaller $k$, but without changing the number of divisors of $M_k$.) Henceforth, we assume that $a_i = 0$ for all $i \ge 4$. All the possibilities are summarised in the following table:

Of course, one can try to be lucky, and simply start to determine the numbers of divisors of $1 \times 20, 2 \times 20, 3 \times 20, \dots$, and soon end up with $12 \times 20$ having $20$ positive divisors (for the first time in this sequence). Hence, $12 \times 20$ is the smallest such multiple.