Ok, I think that I can give you a more sensible answer.
Here is the expression for the Schawrtzschild radius in terms of the Mass:
$r=\frac{2 G M}{c^2}$
Now substitute for the mass in terms of the density (
$M=\frac{4 \pi r^3}{3}$) and get:
$r=\frac{8 \pi G \rho r^3}{3 c^2}$
Now solve for r,
Solve[r == (8 G \[Pi] r^3 \[Rho])/(3 c^2), r]
and get
{{r -> 0},
{r -> -((c Sqrt[3/(2 \[Pi])])/( 2 Sqrt[G] Sqrt[\[Rho]]))},
{r -> (c Sqrt[3/(2 \[Pi])])/( 2 Sqrt[G] Sqrt[\[Rho]])}}
and the last solution is clearly the physical one you are looking for:
$\frac{\sqrt{\frac{3}{2 \pi }} c}{2 \sqrt{G} \sqrt{\rho }}$
Is this the correct interpretation of your question?
I hope this helps...