7. Given a sphere with a radius of 3. Then the maximum volume of the inscribed regular tetrahedron is $\qquad$ .


7. $8 \sqrt{3}$.

Let the radius of the small circle at the base be $r$. Then the volume of the tetrahedron is
$$
\begin{array}{l}
V=\frac{1}{3} \times \frac{3 \sqrt{3}}{4} r^{2}\left(\sqrt{3^{2}-r^{2}}+3\right) . \\
\text { Let } x=\sqrt{9-r^{2}} \text {, we get } \\
V=\frac{\sqrt{3}}{4}\left(9-x^{2}\right)(x+3) . \\
\text { By } 32-\left(9-x^{2}\right)(x+3) \\
=32-9 x-27+x^{3}+3 x^{2} \\
=(x-1)^{2}(x+5) \geqslant 0 \\
\Rightarrow V=\frac{\sqrt{3}}{4}\left(9-x^{2}\right)(x+3) \\
\quad \leqslant \frac{\sqrt{3}}{4} \times 32=8 \sqrt{3},
\end{array}
$$

When and only when $x=1$, the equality holds.
Therefore, the maximum volume of the inscribed regular tetrahedron is $8 \sqrt{3}$.