23. As shown in Figure 5, a sphere is tangent to both the base and the side of a frustum, and the volume of the frustum is twice that of the sphere. Then the ratio of the radius of the lower base to the radius of the upper base of the frustum is ( ).
(A) $\frac{3}{2}$
(B) $\frac{1+\sqrt{5}}{2}$
(C) $\frac{3+\sqrt{5}}{2}$
(D) 2
(E) $\sqrt{3}$
23. C.

Consider the axial section as shown in Figure 9.
Let the radii of the lower and upper bases of the frustum be $R_{1}$ and $R_{2}$ respectively, with $R_{1} > R_{2}$, and the radius of the inscribed sphere be $r$.
Since the volume of the frustum is twice the volume of the sphere, we have:
$$
\begin{array}{l}
2 \times \frac{4}{3} \pi r^{3}=\frac{1}{3} \pi\left(R_{1}^{2}+R_{1} R_{2}+R_{2}^{2}\right) \cdot 2 r \\
\Rightarrow 4 r^{2}=R_{1}^{2}+R_{1} R_{2}+R_{2}^{2} .
\end{array}
$$

Since the sphere $O$ is inscribed in the frustum, we know that $O C \perp A B$.
Thus, $A O$ bisects $\angle A$, and $B O$ bisects $\angle B$.
Also, $A O_{2} \parallel B O_{1}$, so $A O \perp B O$.
Therefore, $r^{2}=R_{1} R_{2}$.
From equations (1) and (2), we have
$$
\left(\frac{R_{1}}{R_{2}}\right)^{2}-3\left(\frac{R_{1}}{R_{2}}\right)+1=0 \text {. }
$$

Given $R_{1}>R_{2}$, solving yields $\frac{R_{1}}{R_{2}}=\frac{3+\sqrt{5}}{2}$.