Initial radius of wire, $r_1 = r$ and
final radius of wire, $r_2 = r/2 = 0.5r$.
Since volume of the wire after stretch remains constant,
therefore $l_1 A_1 = l_2A_2$
or $\frac{l_{1}}{l_{2}}=\frac{A_{2}}{A_{1}}=\frac{r^{2}_{2}}{r^{2}_{1}}$
$=\left(\frac{0.5r}{r}\right)^{2}=\frac{1}{4}.$
The resistance $\left(R\right)=\rho \frac{l}{A}\propto \frac{l}{A}.$
Therefore $\frac{R_{1}}{R_{2}}=\frac{l_{1}}{l_{2}}\times\frac{A_{2}}{A_{1}}$
$=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}$
or $R_{2}=16R_{1}=16R.$