3. Four spheres with a radius of 2 are pairwise tangent. Then the edge length of the circumscribed regular tetrahedron is
A. $2(1+\sqrt{3})$
B. $4(1+\sqrt{3})$
C. $4(1+\sqrt{6})$
D. $2(1+\sqrt{6})$
3. $\mathrm{C}$ The centers of the four balls $O_{1}, O_{2}, O_{3}, O_{4}$ form a regular tetrahedron $O_{1} O_{2} O_{3} O_{4}$ with an edge length of 4. Its centroid $O$ divides the height of the tetrahedron $O_{1} O_{2} O_{3} O_{4}$ in the ratio $3: 1$. Since the height of the tetrahedron is $\sqrt{a^{2}-\left(\frac{\sqrt{3}}{3} a\right)^{2}}=\frac{4 \sqrt{6}}{3}$, the distance $d$ from $O$ to each face of the tetrahedron is $d=\frac{1}{4} \cdot \frac{4 \sqrt{6}}{3}=\frac{\sqrt{6}}{3}$.
The circumscribed tetrahedron $\mathrm{O}^{\prime}{ }_{1} \mathrm{O}_{2}^{\prime} \mathrm{O}^{\prime}{ }_{3} \mathrm{O}^{\prime}{ }_{4}$ of the four balls has each of its faces parallel to the corresponding faces of the tetrahedron $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$, and the distance between them is 2.
Therefore, the tetrahedron $\mathrm{O}_{1}^{\prime} \mathrm{O}_{2}^{\prime} \mathrm{O}^{\prime}{ }_{3} \mathrm{O}^{\prime}{ }_{4}$ is a regular tetrahedron, sharing the same center $O$ with the tetrahedron $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$, and the two tetrahedrons are similar, with the similarity center at $O$ and the similarity ratio $\frac{d}{d+2}=1+\sqrt{6}$. Thus, the edge length of the tetrahedron $O^{\prime}{ }_{1} O^{\prime}{ }_{2} O_{3}^{\prime} O_{4}^{\prime}$ is $4(1+\sqrt{6})$.