Standard wedge - block problem.

Let the angle which the block makes with the horizontal while sliding down the wedge at any instant be $\theta$

Let the block's velocity be 'v'

Using Energy Conservation

$\frac {1}{2}mv^2=mgR\sin \theta\ ..(1)$

Also..as the block is in circular motion

$N-mg\sin \theta=\frac {mv^2}{R} ..(2)$

where $N$ is Normal Force exerted by the wedge on the block

On solving the above to equations..we get

$N=3mg\sin \theta$

Now..from the FBD of the wedge we get

$f=N\cos \theta=\mu N_{1} ..(3)$

where $N_{1}$ is normal force on wedge by the ground

Again from the FBD of wedge..we get

$N_{1}=N\sin \theta+mg ....(4)$

Substituting the value of $N_{1}$ in eqn $3$

$\mu =\frac {N\cos \theta}{mg+N\sin \theta}$

Now putting $N=3mg\sin \theta$

we get $\mu=\frac {\frac {3}{2}\sin 2\theta}{1+3(\sin \theta)^2}$ .. $5$

To minimize $\mu.... \frac {d\mu}{d\theta}=0$

$\Rightarrow 3\cos 2\theta(1+3(\sin \theta)^2)-(\frac {3}{2}\sin 2\theta)(3\sin 2\theta)=0$

Now putting $\cos 2\theta=1-2(\sin\theta)^2$ and $\sin 2\theta=2\sin \theta \cos \theta$

we get

$\sin \theta = \sqrt{\frac {1}{5}}$

Substituting $sin \theta$ in eqn $5$

we get

$\boxed{\mu=0.75}$