A potential V _ 0 is applied across a uniform wire of resistance R...

The formula to calculate the power dissipated in the first case can be written as

${P}_{1} = {V}_{0}^{2}R . . . \left(1\right)$

When the wire is halved, its resistance will also become half the previous value.

Thus, the formula to calculate the power dissipated in the second case can be written as

$\begin{cases}{P}_{2} & = & {V}_{0}^{2}\dfrac{R}{2} \times 2 \\ & = & \dfrac{4{{V}_{0}}^{2}}{R} . . . \left(2\right)\end{cases}$

Divide equation (2) by equation (1) to calculate the required ratio.

$\begin{cases}\dfrac{{P}_{2}}{{P}_{1}} & = & \dfrac{\dfrac{4{{V}_{0}}^{2}}{R}}{\dfrac{{{V}_{0}}^{2}}{R}} \\ & = & 4 \\ & = & \sqrt{16}\end{cases}$

Hence, $x = 16$.