6. A frustum of height 8 contains a sphere $O_{1}$ with a radius of 2, the center $O_{1}$ is on the axis of the frustum, and the sphere $O_{2}$ is tangent to the top base and the side of the frustum. The frustum can also contain another sphere $O_{2}$ with a radius of 3, such that the sphere $O_{2}$ is tangent to the sphere $O_{1}$, the bottom base, and the side of the frustum at only one point. Apart from the sphere $O_{2}$, the maximum number of additional spheres with a radius of 3 that can be placed inside the frustum is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
6. (B).

Draw the axial section of the frustum through $\mathrm{O}_{2}$, as shown in Figure 1. Then, draw a section through $\mathrm{O}_{2}$ perpendicular to the axis of the frustum; the intersection of this section with the axis of the frustum is a circle $O$. From Figure 1, it is easy to find that $O O_{2}=4$.

This problem is equivalent to: On a circle with center $O$ and radius 4, how many more points can be placed, excluding $\mathrm{O}_{2}$, such that the circles with these points and $\mathrm{O}_{2}$ as centers and 3 as the radius have at most one common point. From Figure 2,
$$
\sin 45^{\circ}<\sin \theta=\frac{3}{4} \sin 60^{\circ},
$$

we have $45^{\circ}<\theta<60^{\circ}$.
Therefore, the maximum number of additional points that can be placed is
$$
\left\lceil\frac{360^{\circ}}{20}\right\rceil-1=3-1=2
$$

points, satisfying the above conditions.
Thus, the frustum can contain at most 2 more spheres with a radius of 3.