Step 1: Use the fact that $n\sin\theta$ stays constant across a stack of parallel interfaces.
When light crosses several parallel-sided layers of different media one after another, the quantity $n\sin\theta$ measured from the very first medium is conserved all the way through, because each interface links the medium just before it to the medium just after it in one long chain. For the ray going from air into oil and then into water, $n_{air}\sin\theta_{air} = n_{oil}\sin\theta_{oil} = n_{water}\sin\theta_{water}$.
Step 2: Write down the known values.
$n_{air} = 1$, $\theta_{air} = 30^{\circ}$, $n_{water} = 1.3$. Because the chain rule links air directly to water, we do not need to work out the intermediate angle inside the oil at all.
Step 3: Apply the invariant directly between air and water.
\[ n_{air}\sin\theta_{air} = n_{water}\sin\theta_{water} \] \[ 1 \times \sin 30^{\circ} = 1.3 \times \sin\theta_{water} \]
Step 4: Solve for the angle in water.
\[ \sin\theta_{water} = \frac{\sin 30^{\circ}}{1.3} = \frac{0.5}{1.3} = \frac{1}{2.6} \] So $\theta_{water} = \sin^{-1}\left(\frac{1}{2.6}\right)$.
Step 5: Rule out the other options.
$\sin^{-1}(1/2)$ is the angle if the ray had stayed in air; $\sin^{-1}(1.3/1.5)$ swaps the two refractive indices around; $\sin^{-1}(2/3)$ does not follow from this data at all. Only $\sin^{-1}(1/2.6)$ comes from the conserved quantity $n\sin\theta$.
Final Answer:
\[ \boxed{\sin^{-1}\left(\frac{1}{2.6}\right)\ \text{rad}} \]