The normal reaction between the wedge and the block is

$N^{\text{'}}=mg\cos \theta$

So the net horizontal force on the wedge is

$F=mg\cos \theta \sin \theta$

The normal reaction between the wedge and the ground is

$N=2 mg+mg{\cos }^2\theta$

$N=2\mathrm{ }mg+mg\mathrm{ }\mathrm{c}\mathrm{o}{\mathrm{s}}^2\theta$

Now, $F=\mu N$ gives

$\mu =\frac{F}{N}=\frac{mg\cos \theta \sin \theta }{2 mg+mg{\cos }^2\theta }$

$=\frac{\cos \theta \sin \theta }{2+{\cos }^2\theta }$

For $\theta ={45}^{\mathrm{o}},\mathrm{ }\mathrm{ }\mu =0.2$