6. Let the inradius of a regular tetrahedron be 1. Then the minimum value of its volume is $\qquad$ .


$6.8 \sqrt{3}$.
Let the center of the base of the regular tetrahedron $P-ABC$ be $H$. Then the center of the inscribed sphere $O$ lies on $PH$.

If $AH$ intersects $BC$ at point $D$, then $PD \perp BC$, and the point of tangency $E$ of the sphere $O$ with the plane $PBC$ lies on $PD$.
Considering the section
$\triangle PAD$.
Let $\angle PDA = 2\theta, PH = h$, and the side length of the equilateral $\triangle ABC$ be $a$. Then
$$
\begin{array}{c}
AD = \frac{\sqrt{3}}{2} a, \\
DH = DE = \frac{\sqrt{3}}{6} a .
\end{array}
$$

Since $DH = \cot \theta$, then
$$
\begin{array}{l}
a = 2 \sqrt{3} DH = 2 \sqrt{3} \cot \theta, \\
h = DH \tan 2\theta = \frac{2}{1 - \tan^2 \theta} .
\end{array}
$$

Thus $V = \frac{1}{3} \times \frac{\sqrt{3}}{4} a^2 h$
$$
= 2 \sqrt{3} \times \frac{1}{\tan^2 \theta (1 - \tan^2 \theta)} \geqslant 8 \sqrt{3} .
$$