3. The sum of the upper base radius $r$ and the lower base radius $R (R > r)$ of a frustum is 6 times the slant height $l$, and the areas of the upper base, lateral surface, and lower base form a geometric sequence. The height of this frustum is $20 - 3r^2$, then the maximum volume of the frustum is $\qquad$
$$
\begin{array}{l}
\text { 3. } \frac{500}{9} \pi(2+\sqrt{3}) . \\
r+R=6 l,(\pi(r+R) l)^{2}=\pi r^{2} \cdot \pi R^{2}, \\
l=\frac{R r}{R+r},(r+R)^{2}=6 R r .
\end{array}
$$

Given $R>r$, then $R=(2+\sqrt{3}) r$.
$$
\begin{array}{l}
V=\frac{1}{3} \pi\left(20-3 r^{2}\right)\left(\pi^{2}-1-R r+r^{2}\right) \\
=\frac{1}{3} \pi\left(20-3 r^{2}\right)(10+5 \sqrt{3}) r^{2} \\
=\frac{5}{9} \pi(2+\sqrt{3})\left(20-3 r^{2}\right) \cdot 3 r^{2} \\
\leqslant \frac{5}{9} \pi(2+\sqrt{3})\left(\frac{20}{2}\right)^{2}
\end{array}
$$