A tetrahedron with face area $F$ and altitude (to that face) $a$ has volume $\frac{1}{3}a F$.
Now, consider a sub-tetrahedron determined by a given face, with fourth vertex at the center of the inscribed sphere. Because the sphere is tangent to the face, the altitude to that face is a radius, say, of length $r$. If the big tetrahedron is regular (with face area $F$ and altitude $a$), then the four sub-tetrahedra are congruent, with volume $\frac{1}{3}r F$.
Together, the sub-tetrahedra fill the big tetrahedron, so the four sub-volumes add up to the big volume:
$4 \cdot \frac{1}{3} r F = \frac{1}{3} a F$
so that $4 r = a$. You're given that $a=36$; consequently, $r = 36/4 = 9$.
Note. Regular or not, the basic argument shows that the volume ($V$) of a tetrahedron is determined by its inradius ($r$) and its total surface area ($S$):
$V = \frac{1}{3} r S$
This is analogous to the fact that the area ($A$) of a triangle is given by the inradius ($r$) and its perimeter ($p$):
$A = \frac{1}{2}r p$
The pattern continues into $d$-dimensional space, with the multiplied constant being $\frac{1}{d}$.