6. If the volume of the circumscribed sphere of a regular tetrahedron is $36 \pi$, then the volume of the regular tetrahedron is $\qquad$ .
6. $8 \sqrt{3}$.

To complete the regular tetrahedron $A B_{1} C D_{1}$ into a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, its circumscribed sphere is precisely the circumscribed sphere of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, with the diameter being $A C_{1}$. It is easy to find that the radius of the circumscribed sphere is 3.
Thus, $A C_{1}=6, A A_{1}=2 \sqrt{3}$.
Therefore, the volume of the required regular tetrahedron is
$$
\begin{array}{l}
V_{\text {cube }}-4 V_{\text {tetrahedron } A A_{1} B_{1} D_{1}} \\
=(2 \sqrt{3})^{3}-4 \times \frac{1}{3} \times \frac{1}{2}(2 \sqrt{3})^{3}=8 \sqrt{3} .
\end{array}
$$