At point $A$ , by Snell's law



$\mu =\frac{\sin {45}^o}{\sin r}\Rightarrow \sin r=\frac{1}{\mu \sqrt{2}}$ ... (i)

At point $B$ , total internal reflection with $\sin i_1=\frac{1}{\mu }$

From figure, $i_1={90}^o-r$ .... (ii)

$\sin \left({90}^o-r\right)=\frac{1}{\mu }\Rightarrow \cos r=\frac{1}{\mu }$ .... (iii)

Now, $\cos r=\sqrt{1-{\sin }^{2}r}=\sqrt{1-\frac{1}{2{\mu }^2}}=\sqrt{\frac{2{\mu }^2-1}{{\mu }^2}}$

From equation (ii) and (iii), $\frac{1}{\mu }=\sqrt{\frac{2{\mu }^2-1}{2{\mu }^2}}$

Squaring both sides and then solving, we get

$\mu =\sqrt{\frac{3}{2}}$