1. Use the invariant $n\sin\phi$ across successive parallel interfaces
   Since the boundaries between Regions I, II, III and IV are parallel, Snell’s law at each interface gives $n\sin\phi=\text{constant}$ where $\phi$ is the angle the ray makes with the normal in that region.

2. Condition for ‘just misses entering Region IV’
   This means that at the interface between Region III and Region IV, the refracted ray in Region IV is just not formed. So the incidence angle in Region III is the critical angle for the III–IV boundary.

   Hence, $\sin\phi_3=\frac{n_4}{n_3}$

   Given: $n_3=\frac{n_0}{6},\qquad n_4=\frac{n_0}{8}$ therefore $\sin\phi_3=\frac{\frac{n_0}{8}}{\frac{n_0}{6}}=\frac{6}{8}=\frac{3}{4}$

3. Relate the angle in Region I to the angle in Region III
   Because $n\sin\phi$ remains constant across the parallel layers, $n_1\sin\theta=n_3\sin\phi_3$

   Here, $n_1=n_0,\qquad n_3=\frac{n_0}{6},\qquad \sin\phi_3=\frac{3}{4}$

   So, $n_0\sin\theta=\frac{n_0}{6}\cdot\frac{3}{4}$ $\sin\theta=\frac{1}{6}\cdot\frac{3}{4}=\frac{1}{8}$

4. Find $\theta$ $\theta=\sin^{-1}\left(\frac{1}{8}\right)$

5. Check options
   This matches Option B.

Final Answer: $\boxed{\theta=\sin^{-1}\left(\frac{1}{8}\right)}$