The length of a given cylindrical wire is increased to double of...

As volume is constant, so, length doubled and area reduces to half.

Resistance of a wire is given by $R = \dfrac{ρl}{A}$, where, $ρ$ is resistivity of a material.

Here, ${R}_{1} = \dfrac{ρl}{A}$ and ${R}_{2} = \dfrac{ρ\left(2l\right)}{\dfrac{A}{2}} = 4R$

So percentage increase in the resistance will be $= \dfrac{4R - R}{R} \times 100$