Let $r_\min$ be the radius of the minimal enclosing sphere.
The circumradius is an upper bound on $r_\min$, because the circumsphere encloses the tetrahedron.
$\sqrt{3/8}$ times the length of the longest edge is an upper bound on $r_\min$, because for a given length $\ell$, the tetrahedron with the largest minimal enclosing sphere and no edge longer than $\ell$ is the regular tetrahedron of edge length $\ell$, and its circumradius is $\sqrt{3/8}\ell$.
Putting that together, the value $\min(r_{\text{circ}},\sqrt{3/8}\ell_\max)$ is an upper bound on $r_\min$.
I don't know how the inradius can be used to improve the bound. But I'm pretty sure all you can get are bounds; specifying the circumradius, length of longest edge, and inradius is not sufficient to uniquely specify a tetrahedron.