The radius $r=$ $\qquad$ of the exsphere (i.e., the sphere that is tangent to one face and the extensions of the other three faces) of a regular tetrahedron with edge length 1.
$4 \frac{\sqrt{6}}{6}$. Let the regular tetrahedron be $ABCD$, the exsphere is tangent to the face $BCD$, the radius of the sphere is $r$, and the center of the sphere is $P$, then
that is
$$
\left(\frac{1}{3} \cdot r \cdot \frac{1}{2} \cdot 1^{2} \cdot \frac{\sqrt{3}}{2}\right) \cdot(3-1)=\frac{\sqrt{2}}{12} \text {, }
$$

so $r=\frac{\sqrt{6}}{6}$.