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 \(P_2=\frac{V_0^2}{R^{\prime}} \times 2 \Rightarrow \frac{V_0^2}{R} \times 4\)

\(\frac{P_2}{P_1}=4=\frac{\sqrt{x}}{1}\)

\(x=16\)