Eight copper wire of length l and diameter d are joined in parallel...

Each wire has resistance, $r = ρ\dfrac{4l}{\pi{d}^{2}}$

Eight wire in parallel, then equivalent resistance is

$R = \dfrac{r}{8} = \dfrac{ρl}{2\pi{d}^{2}}$

Single copper wire of length $2l$ has resistance $R = ρ\dfrac{2l \times 4}{\pi{d}_{1}^{2}}$.

Comparing both equations, we get,

$ρ\dfrac{8l}{\pi{d}_{1}^{2}} = \dfrac{ρl}{2\pi{d}^{2}}$

$⇒ {d}_{1} = 4d$