c
$P=V I=I^2 R=\frac{V^2}{R}$
Now $R=\frac{\rho l}{A}$
If wire is cut in two equal half
$R^{\prime}=\frac{R}{2}$
Initial $P_1=\frac{V_0^2}{R}$
After $P_2=\frac{V_0^2}{R^{\prime}} \times 2 \Rightarrow \frac{V_0^2}{R} \times 4$
$\frac{P_2}{P_1}=4=\frac{\sqrt{x}}{1}$
$x=16$