Answer
$\mu = 0.43$
Work Step by Step
Let $d$ be the length of the ramp.
We can find the acceleration when the slide is frictionless:
$F = ma$
$mg~sin~\theta = ma$
$a = \frac{mg~sin~\theta}{m}$
$a = g~sin~\theta$
We can write an expression for the time required to slide down a frictionless ramp:
$d = \frac{1}{2}at_1^2$
$t_1 = \sqrt{\frac{2d}{a}}$
$t_1 = \sqrt{\frac{2d}{g~sin~\theta}}$
We can find the acceleration when there is friction:
$mg~sin~\theta - mg~\mu~cos~\theta = ma$
$a = g~sin~\theta - g~\mu~cos~\theta$
We can write an expression for the time required to slide down the ramp when there is friction:
$d = \frac{1}{2}at_2^2$
$t_2 = \sqrt{\frac{2d}{a}}$
$t_2 = \sqrt{\frac{2d}{g~sin~\theta - g~\mu~cos~\theta}}$
It is given in the question that $t_2 = 2t_1$
We can find $\mu$:
$t_2 = 2t_1$
$\sqrt{\frac{2d}{g~sin~\theta - g~\mu~cos~\theta}} = 2\sqrt{\frac{2d}{g~sin~\theta}}$
$\frac{2d}{g~sin~\theta - g~\mu~cos~\theta} = \frac{8d}{g~sin~\theta}$
$4g~sin~\theta - 4g~\mu~cos~\theta = g~sin~\theta$
$4\mu~cos~\theta = 3~sin~\theta$
$\mu = 0.75~tan~\theta$
$\mu = 0.75~tan~35^{\circ}$
$\mu = 0.43$