#### Answer

We need to increase the height of the track by a factor of 4

#### Work Step by Step

The potential energy at the top of the track is equal to the kinetic energy at the bottom of the track. Let the height be $h_0$ when the speed at the bottom is $v_0$.
$PE_0 = KE_0$
$mg~h_0 = \frac{1}{2}mv_0^2$
$h_0 = \frac{v_0^2}{2g}$
Suppose the speed at the bottom is $2v_0$. We can find an expression for the initial height $h_2$.
$PE_2 = KE_2$
$mg~h_2 = \frac{1}{2}m(2v_0)^2$
$h_2 = 4\times~\frac{v_0^2}{2g}$
$h_2 = 4\times h_0$
We need to increase the height of the track by a factor of 4.