Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
From the geometry of the given figure, we can see that the first refracted angle inside the prism is
$$\theta =90^\circ -\gamma\tag 1$$
where, from the isosceles triangle,
$$2\gamma+\alpha=180^\circ$$
So that
$$\gamma=\dfrac{180^\circ-\alpha}{2}$$
Plugging into (1);
$$\theta =90^\circ -\dfrac{180^\circ-\alpha}{2}=90^\circ-90^\circ+\dfrac{\alpha}{2}$$
Thus,
$$\theta=\dfrac{\alpha}{2}\tag2$$
Applying Snell's law,
$$n_{air}\sin\beta=n\sin\theta=n\sin\left(\frac{\alpha}{2}\right)$$
where $n_{air}=1$, thus
$$\boxed{\beta=\sin^{-1}\left[ n\sin\left(\frac{\alpha}{2} \right) \right]}$$
$$\color{blue}{\bf [b]}$$
Solving the boxed formula above for $n$, and the prism is an equilateral triangle, so $\alpha=60^\circ$.
$$n=\dfrac{ \sin\beta}{\sin\left(\dfrac{\alpha}{2}\right)} $$
Plugging the known;
$$n=\dfrac{ \sin52.2^\circ}{\sin\left(\dfrac{60^\circ}{2}\right)}=\color{red}{\bf1.58} $$
