on a rough surface acceleration of a body down a plane

$a=(g \;sin \theta-\mu g \cos \theta)$

'2t' is time taken to go down a rough inclined plane of length 's'

't' is time taken to go down a smooth inclined plane of length 's'

$s=\large\frac{1}{2}$$ [g \sin \theta -\mu g \cos \theta](2t)^2$-----(1)

On a smooth surface $a=g \sin \theta$

$s= \large\frac{1}{2} $$(g \sin \theta)t^2$-----(2)

Substituting (2) in (1) and solving

$\mu =1- \large\frac{1}{2^2}$

$\mu =\large \frac{3}{4}$

$\qquad=0.75$

Hence b is the correct answer.