$$ \mathrm{R}_{\text {initial }}=\frac{\rho \ell}{A}=5 \Omega $$

$\because$ Volume of wire is constant in stretching

$$ \begin{aligned} & \mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{f}} \\\\ & \mathrm{A}_{\mathrm{i}} \ell_{\mathrm{i}}=\mathrm{A}_{\mathrm{f}} \ell_{\mathrm{f}} \\\\ & \mathrm{A} \ell=\mathrm{A}^{\prime}(5 \ell) \\\\ & \mathrm{A}^{\prime}=\frac{\mathrm{A}}{5} \\\\ & \mathrm{R}_{\mathrm{f}}=\frac{\rho \ell_{\mathrm{f}}}{\mathrm{A}_{\mathrm{f}}}=\frac{\rho(5 \ell)}{\left(\frac{\mathrm{A}}{5}\right)} \\\\ & =25\left(\frac{\rho \ell}{\mathrm{A}}\right) \\\\ & =25 \times 5=125 \Omega \end{aligned} $$