$A$ piece of wire is redrawn,without change in volume,so that its radius is halved. Compare the new resistance with the original resistance.

(D) Let the length of the original wire be $l$ and its radius be $r$.
The original resistance is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.
The original volume is $V_1 = \pi r^2 l$.
When the radius is halved,the new radius becomes $r' = \frac{r}{2}$. Let the new length be $l'$.
Since the volume remains constant,$V_1 = V_2$,so $\pi r^2 l = \pi (\frac{r}{2})^2 l'$.
$\pi r^2 l = \pi \frac{r^2}{4} l'$,which simplifies to $l' = 4l$.
The new resistance is $R' = \rho \frac{l'}{\pi (r')^2} = \rho \frac{4l}{\pi (r/2)^2}$.
$R' = \rho \frac{4l}{\pi (r^2/4)} = 16 \left( \rho \frac{l}{\pi r^2} \right)$.
Therefore,$R' = 16R$.
The new resistance is $16$ times the original resistance.