On the rough incline, $a_{1}=g(\sin \theta-\mu \cos \theta)$

$t_{1}=$ time taken

On the frictionless incline, $a_{2}=g \sin \theta$

$t_{2}=$ time taken, $t_{1}=2 t_{2}$

$s=u t+\frac{1}{2} a t^{2}$

$s_{1}=0+\frac{1}{2} g(\sin \theta-\mu \cos \theta) t_{1}^{2}$

$s_{2}=0+\frac{1}{2} g \sin \theta t_{2}^{2}$

As $s_{1}=s_{2}$

$\frac{1}{2} g(\sin \theta-\mu \cos \theta) t_{1}^{2}=\frac{1}{2} g \sin \theta t_{2}^{2}$

$\frac{\sin \theta-\mu \cos \theta}{\sin \theta}=\frac{t_{2}^{2}}{t_{1}^{2}}$

$\Rightarrow 1-\mu \cot \theta=\frac{t_{2}^{2}}{\left(2 t_{2}\right)^{2}}$

$\Rightarrow 1-\mu \cot \theta=\frac{1}{4}$

$\Rightarrow \mu \cot \theta=1-\frac{1}{4}=\frac{3}{4}$

$\therefore \mu=\frac{3}{4 \cot \theta}$