The specific resistance of a wire is $\rho$, its volume is 3 m³ and its resistance is 3$\Omega$, then its length will be

Volume of wire =Al

Given, 3=Al

$$\begin{gathered} \therefore A=\frac{3}{l} \\ Now, resis\tan ce, R =\rho \frac{l}{A} \\ 3=\frac{\rho l}{\left(3/l\right)}=\frac{\rho l^2}{3} \\ \therefore l^2=\frac{9}{\rho } \\ \therefore l=\frac{3}{\sqrt{\rho }} \end{gathered}$$