A prism with angles of 45°, 45° and 90° is made of glass with refractive index 1.50…

Step 1. In air. Total reflection needs $\sin 45^\circ > \dfrac{1}{n}$, so $n > \dfrac{1}{\sin 45^\circ} = \sqrt{2} = 1.41$. The glass has $1.50$, so it works.

Step 2. The critical angle in water. $\sin C = \dfrac{1.33}{1.50} = 0.8867$, so $C = 62.5^\circ$.

Step 3. The ray in the water. The angle of incidence, $45^\circ$, is less than $62.5^\circ$, so most of the light leaves the glass. Snell's law gives $\sin\theta = \dfrac{1.50}{1.33}\sin 45^\circ = 0.7975$, so $\theta = 52.9^\circ$ with the normal.

Step 4. The index needed in water. We need $\sin 45^\circ > \dfrac{1.33}{n}$, so $n > 1.33 \times \sqrt{2} = 1.88$.