wire $=P=\frac{V^{2}}{R}$

Resistance of the wire of length, $L$

$R_{1}=\frac{\rho L}{A}=\frac{\rho L}{\pi r^{2}}$

$\therefore$ Power, $P_{1}=\frac{V^{2}}{R_{1}}$

Resistance of the wire when length is halved

i.e., $L / 2$

$R_{2}=\frac{\rho \frac{L}{2}}{\pi(2 r)^{2}}=\frac{\rho L}{\pi 8 r^{2}}=\frac{R_{1}}{8}$

$\therefore$ Power, $P_{2}=\frac{V}{\frac{R_{1}}{8}}=\frac{8 V}{R_{1}}$

or, $\mathrm{P}_{2}=8 \mathrm{P}_{1}$ i.e., power increased $8$ times of previous or original wire.