6. Given that the radius of the inscribed sphere of a regular tetrahedron is 1, then the volume of the regular tetrahedron is
Given $8 \sqrt{3}$.
Analysis: Let the edge length of the regular tetrahedron be $a$, then the volume of the regular tetrahedron is $\frac{1}{3} \times \frac{\sqrt{3}}{4} \times a^{2} \times \frac{\sqrt{6}}{3} a$, and the total surface area is $\frac{\sqrt{3}}{4} a^{2} \times 4$, so
$$
\frac{1}{3} \times \frac{\sqrt{3}}{4} \times a^{2} \times 1 \times 4=\frac{1}{3} \times \frac{\sqrt{3}}{4} \times a^{2} \times \frac{\sqrt{6}}{3} a
$$

Solving for $a$ gives $a=2 \sqrt{6}$, thus the volume of the regular tetrahedron is
$$
\frac{1}{3} \times \frac{\sqrt{3}}{4} \times a^{2} \times \frac{\sqrt{6}}{3} a=8 \sqrt{3}
$$