Solution:

The answer is that $A$ is larger. First,
$$\begin{aligned} \left(\frac{19}{20}\right)^{20} & = 0.95^{20} = 0.9025^{10} > 0.9^{10} \\ & = 0.81^{5} = 0.8^{5} = 0.32768 > 0.05 = \frac{1}{20} \end{aligned}$$
which implies that $19^{20} > 20^{19}$. Thus, $A = 20^{19^{20}} > 20^{20^{19}} > 19^{20^{19}} = B$, as desired.