Solve the external figure. Let the radii of the upper and lower bases of the frustum be $\mathrm{r}$ and $\mathrm{R}$, and the height be $\mathrm{h}$, then
$$\begin{array}{l} \frac{1}{3} \pi h\left(r^{2}+r R+R^{2}\right) \\ =2 \cdot \frac{1}{3} \pi \hbar^{2} h . \end{array}$$

Simplifying and rearranging, we get $R^{2}-r R-r^{2}=0$.
Solving it, $R=\frac{(1+\sqrt{5}) \mathrm{r}}{2}$ and $R=\frac{(1-\sqrt{5}) \mathrm{r}}{2}$ (not applicable, discard).
Therefore, $\frac{r}{R}=\frac{2}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{2}$.