
$\begin{aligned}
& \mathrm{L}=25 \mathrm{m}, \mathrm{A}=5 \mathrm{mm}^2=5 \times 10^{-6} \mathrm{m}^2 \
& \rho=2 \times 10^{-6} \Omega \mathrm{m} \
& \mathrm{R}_{\text {wire }}=\frac{\rho \mathrm{L}}{\mathrm{A}}=\frac{2 \times 10^{-6} \times 25}{5 \times 10^{-6}}=10 \
& \mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}}{4}=\frac{10}{4}=2.5 \Omega
\end{aligned}$
Answer does not match with NTA option.