The setup is

(1) $\displaystyle W = \frac{kd^2}{l} $

where k is the constant of variation.

The percentage increase of d is $\displaystyle \frac{x}{100} \cdot d $ W remains the same but l becomes 2l.

$\displaystyle W = \frac{k(d + \frac{xd}{100})^2}{2l} $

(2) $\displaystyle W = \frac{kd^2(1 + \frac{x}{100})^2}{2l} $

Set (1) = (2) and solve for x

$\displaystyle \frac{kd^2}{l} = \frac{kd^2(1 + \frac{x}{100})^2}{2l} $

We see k and l and d^2 cancel immediately giving

$\displaystyle 2 = (1 + \frac{x}{100})^2 $

Can you continue from here?

I'm already estimating x about 41 %...