AMC 12 Mock Exam 1 Problem 14

A regular tetrahedron of edge $6$ can be inscribed in a cube of edge $\frac{6}{\sqrt2}=3\sqrt2$ by taking four alternate vertices of the cube; its six edges are face diagonals of the cube, one on each face. The center of the tetrahedron is the center of the cube.

The distance from the cube's center to a face diagonal: the diagonal passes through the center of its face, and the segment from the cube's center to that face center is perpendicular to the face, hence perpendicular to the diagonal. So the distance is half the cube's edge, $r=\frac{3\sqrt2}{2}$, and the same holds for all six edges. The sphere has surface area $4\pi r^2=4\pi\cdot\frac{18}{4}=\boxed{18\pi}$.