10. Given a regular tetrahedron $S-ABC$ with side length of the lateral edges being $6 \sqrt{3}$, and the side length of the base being 6. Then the surface area of the circumscribed sphere of this regular tetrahedron is $\qquad$
10. $\frac{243 \pi}{3}$.

As shown in Figure 2, draw $S E \perp$ plane $A B C$ at point $E$, and let the center of the sphere be $O$.

In the regular tetrahedron $S-$ $A B C$, by the problem statement,
$$
\begin{array}{l}
B E=\frac{2}{3} \times \frac{\sqrt{3}}{2} \times 6 \\
=2 \sqrt{3}, \\
S E=\sqrt{S B^{2}-B E^{2}}=4 \sqrt{6} .
\end{array}
$$

Since the distance from the center of the sphere $O$ to the four vertices is equal (all equal to the radius $R$ of the circumscribed sphere of the regular tetrahedron), we have
$$
O B=R, O E=4 \sqrt{6}-R \text {. }
$$

In the right triangle $\triangle B O E$, $O B^{2}=B E^{2}+O E^{2}$, that is,
$$
R^{2}=12+(4 \sqrt{6}-R)^{2} \Rightarrow R=\frac{9 \sqrt{6}}{4} \text {. }
$$

Therefore, the required value is $4 \pi R^{2}=\frac{243 \pi}{3}$.