6. If a sphere of radius $R$ contains four identical smaller spheres of radius $r$, then the maximum possible value of $r$ is $(\quad)$.
(A) $\frac{\sqrt{3}}{2+\sqrt{3}} R$
(B) $\frac{\sqrt{6}}{3+\sqrt{6}} R$
(C) $\frac{1}{1+\sqrt{3}} R$
(D) $\frac{\sqrt{5}}{2+\sqrt{5}} R$
6. B.

As shown in Figure 2, connecting the centers of the four small balls $O_{1}$, $O_{2}$, $O_{3}$, and $O_{4}$ forms a tetrahedron. Let the projection of point $O_{1}$ on the base $O_{2}O_{3}O_{4}$ be $H$.

Then the center of the large ball $O$ lies on $O_{1}H$, and
$$
\begin{array}{l}
O_{2}H=\frac{\sqrt{3}}{3} \times 2r, \\
O_{1}H=\sqrt{(2r)^{2}-\left(\frac{2\sqrt{3}}{3}r\right)^{2}}=\frac{2\sqrt{6}}{3}r .
\end{array}
$$

Therefore, $O_{1}O=\frac{3}{4}O_{1}H=\frac{\sqrt{6}}{2}r$.
$$
\text{By } \frac{\sqrt{6}}{2}r + r = R \Rightarrow r = \frac{\sqrt{6}}{3+\sqrt{6}}R \text{.}
$$