Eight copper wire of length l and diameter d are joined in parallel to form a single composite conductor of…

Each wire has resistance, $r=\rho \frac{4l}{\pi {\mathrm{d}}^2}$

Eight wire in parallel, then equivalent resistance is

$R=\frac{r}{8}=\frac{\rho l}{2\pi d^2}$

Single copper wire of length $2l$ has resistance $R=\rho \frac{2l\times 4}{\pi d_1^2}$.

Comparing both equations, we get,

$\rho \frac{8l}{\pi d_1^2}=\frac{\rho l}{2\pi d^2}$

$\Rightarrow d_1=4d$