The inductance L of a solenoid of length /, whose windings of mass $m$ are made of material of density D and resistivity $\rho$, is (the winding resistance is R)

For a solenoid, $L={\mu }_0{N}^2\frac{A}{l}$. 

If x is the length of the straight wire and a is the area of cross-section, then

$R=\frac{\rho x}{a} and m=axD$

$Rm=\frac{\rho x}{a}axD, x=\sqrt{\frac{Rm}{\rho D}}$

$Also, x=2 \pi rN, N=\frac{x}{2\pi r} \left(\therefore L=\frac{{\mu }_0{N}^{2}A}{l}\right)$

$\therefore L={\mu }_0{\left(\frac{x}{2\pi r}\right)}^2=\frac{\pi r^2}{l}=\frac{{\mu }_0}{4\pi l}\frac{Rm}{\rho D}$