Answer : $\mu=2\tan\theta$
The acceleration of the block while it is sliding down the upper half of the inclined plane is $g\sin \theta$.
If $\mu$ is the coefficient of kinetic friction between the block and the lower half of the plane,the retardation of the block while it is sliding down the lower half = $-(g\sin \theta-\mu\cos\theta)$
For the block to come to rest at the bottom of the inclined plane,the acceleration in the first half must be equal to the retardation in the second half,i.e.
$g\sin \theta =-(g\sin \theta-\mu g\cos\theta)$
$\mu \cos \theta = 2\sin \theta$
Hence the correct answer is $\mu=2\tan\theta$