$165 \frac{4}{5}$. Let the radius of the small circle be $r$. From Figure 16, it is easy to see that
$$
(1+r)^{2}=(1-r)^{2}+1
$$

So $r=\frac{1}{4}$.
Let $\mathrm{O}_{2} \mathrm{O}_{3}$ intersect the cylindrical surface at $A$ and $B$, then $A B$ is the major axis $2 a$. And
$$
\frac{A C}{A D}=\frac{O_{3} C}{O_{2} D}=\frac{1}{4},
$$

So $A C=\frac{1}{3} . A O_{3}=\sqrt{\left(\frac{1}{3}\right)^{2}+\left(\frac{1}{4}\right)^{2}}=\frac{5}{12} \cdot \frac{\frac{5}{12}}{2 a}=\frac{a_{3} C}{B E}=\frac{\frac{1}{4}}{2}$, so $a=\frac{5}{3}$.
Let the minor axis be $C D=2 b . C D$ clearly passes through the center of the sphere $O_{2}$ and is bisected by $O_{2}$. The shortest distance from $O_{2}$ to the cylindrical surface is 1, so the minimum value of $b$ is 1 (when $C D$ is perpendicular to the plane $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$).
Combining the above, the eccentricity of the ellipse $e=\frac{c}{a}=\frac{\sqrt{a^{2}-b^{2}}}{a} \leqslant \frac{\sqrt{\left(\frac{5}{3}\right)^{2}-1^{2}}}{\frac{5}{3}}=\frac{4}{5}$, so the maximum value of the eccentricity $e$ is $\frac{4}{5}$.
165 Two spheres with a radius of 1 are externally tangent to each other, and both are internally tangent to a cylindrical surface with a radius of 1. Another smaller sphere is externally tangent to both of these spheres and internally tangent to the cylindrical surface. A plane passing through the center of the smaller sphere and the center of one of the larger spheres intersects the cylindrical surface to form an ellipse. The maximum possible value of the eccentricity \( e \) of this ellipse is