Given : Initial radius of wire $\left(r_{1}\right)=r$ ; Initial resistance $=R$ and final radius of wire $\left(r_{2}\right)=$ $0.5 r$. Let $L_{1}$ and $A_{1}$ be the length and cross-sectional area of original wire and $L_{2}, A_{2}$ be the values for the stretched wire
Theretore, $L_{1} \times \pi r_{1}^{2}=L_{2} \times \pi r_{2}^{2}$ or
$L_{1} r=L_{2} \times(0.5 r)^{2}$ or $\frac{L_{1}}{L_{2}}=\frac{1}{4}$ and $\frac{A_{1}}{A_{2}}=\frac{\pi(r)^{2}}{\pi(0.5 r)^{2}}=4$.
Resistance $(R)=\rho \frac{L}{A} \propto \frac{L}{A}$ or
$\frac{R_{1}}{R_{2}}=\frac{L_{1}}{L_{2}} \times \frac{A_{2}}{A_{1}}=\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}$ or $R_{2}=16 R_{1}$