As shown in the diagram, when the incident ray is parallel to base...

From the geometry of the prism, the angle of the prism is $A = 180^\circ - (45^\circ + \theta)$.
The incident ray is parallel to the base, so the angle of incidence $i$ at the first surface is equal to the base angle $45^\circ$. Thus, $i = 45^\circ$.
Applying Snell's law at the first surface: $1 \cdot \sin(45^\circ) = \mu \sin(r_1)$.
Given $\mu = \sqrt{2}$, we have $\frac{1}{\sqrt{2}} = \sqrt{2} \sin(r_1) \implies \sin(r_1) = \frac{1}{2} \implies r_1 = 30^\circ$.
At the second surface, the emergent ray grazes the surface, so the angle of emergence $e = 90^\circ$.
Applying Snell's law at the second surface: $\mu \sin(r_2) = 1 \cdot \sin(90^\circ)$.
$\sqrt{2} \sin(r_2) = 1 \implies \sin(r_2) = \frac{1}{\sqrt{2}} \implies r_2 = 45^\circ$.
The relation between prism angle and internal angles is $A = r_1 + r_2$.
$A = 30^\circ + 45^\circ = 75^\circ$.
Substituting $A$ in the first equation: $75^\circ = 180^\circ - (45^\circ + \theta)$.
$75^\circ = 135^\circ - \theta \implies \theta = 135^\circ - 75^\circ = 60^\circ$.