Step 1: The set up.
Light goes from glass into water. Glass has refractive index $1.5$ and water $1.33$. Glass is the denser medium and water is the rarer one. We need the critical angle, the angle beyond which light cannot leave the glass.
Step 2: The critical angle rule.
When light passes from a denser medium to a rarer one, the critical angle $\theta_c$ obeys
\[ \sin\theta_c = \frac{\mu_{\text{rarer}}}{\mu_{\text{denser}}} \]
Here the rarer medium is water and the denser is glass.
Step 3: Use neat fractions.
Write the indices as simple fractions:
\[ \mu_{\text{glass}} = 1.5 = \frac{3}{2}, \qquad \mu_{\text{water}} = 1.33 \approx \frac{4}{3} \]
Step 4: Substitute.
\[ \sin\theta_c = \frac{\mu_{\text{water}}}{\mu_{\text{glass}}} = \frac{\,4/3\,}{\,3/2\,} \]
Step 5: Simplify the stacked fraction.
Dividing by $3/2$ is the same as multiplying by $2/3$:
\[ \sin\theta_c = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9} \]
Step 6: Write the angle.
\[ \theta_c = \sin^{-1}\!\left(\frac{8}{9}\right) \]
This is option (3).
\[ \boxed{\theta_c = \sin^{-1}\!\left(\frac{8}{9}\right)} \]