2. The interior of a sphere with radius $R$ contains four smaller spheres, each with radius $r$. The maximum possible value of $r$ is ( ).
(A) $\frac{\sqrt{3}}{2+\sqrt{3}} R$
(B) $\frac{\sqrt{6}}{3+\sqrt{6}} R$
(C) $\frac{1}{1+\sqrt{3}} R$
(D) $\frac{\sqrt{5}}{2+\sqrt{5}} R$
2. B.

From the conditions, we know that the centers of the four small balls form the four vertices of a regular tetrahedron, with the center of the large ball being the center of this tetrahedron.

The inradius of a regular tetrahedron can be found using the volume method, which gives $\frac{\sqrt{6}}{2} r$. Therefore,
$$
R=\frac{\sqrt{6}}{2} r+r \Rightarrow r=\frac{\sqrt{6}}{3+\sqrt{6}} R .
$$