3. A circular cone has a height of 12 and a base radius of 5. Let the radius of the inscribed sphere be $r$, and the radius of the circumscribed sphere be $R$. Then $\frac{r}{R}=(\quad)$.
(A) $\frac{89}{169}$
(B) $\frac{83}{169}$
(C) $\frac{80}{169}$
(D) $\frac{60}{169}$
3. C.

Figure 2 is a cross-section of the inscribed sphere in a cone, $O$ is the center of the sphere, $E$ is the point of tangency.
By the similarity ratio, we have
$$
\frac{r}{12-r}=\frac{5}{13} \Rightarrow r=\frac{10}{3} \text {. }
$$

Figure 3 is a cross-section of the circumscribed sphere around the cone, $O$ is the center of the sphere.
By the Pythagorean theorem, we have
$$
\begin{array}{l}
(12-R)^{2}+5^{2}=R^{2} \\
\Rightarrow R=\frac{169}{24} . \\
\text { Hence } \frac{r}{R}=\frac{80}{169} .
\end{array}
$$