Eight wire in parallel, then equivalent resistance is \(\frac{ r }{8}=\frac{\rho \ell}{2 \pi d ^{2}}\)

Single copper wire of length \(2 l\) has resistance

\(R =\rho \frac{2 \ell \times 4}{\pi d _{1}^{2}}=\frac{\rho \ell}{2 \pi d ^{2}}\)

\(d _{1}=4 d\)