c

For solenoid, if  N is the number  of turns, A is cross  section area of solenoid, l is the length of wire then, Self Inductance is $L = \frac{\mu_0 N^2 A}{l_0}$

$A = \pi r^2 , \text{Length of the wire used} l'= 2\pi r \times N$

$\Rightarrow L = \frac{\mu_0 N^2 \pi r^2}{l_0} = \frac{\mu_0 (2\pi rN)^2}{4\pi l} = \frac{\mu_0 l'^2}{4\pi l_0}$

$4\pi L l_0 = \mu_0 l'^2$

$\Rightarrow l'= \sqrt {\frac{4\pi Ll_0}{\mu_0}}$