Refraction of light but slightly twisted This is the question:

(I haven't bothered to type it because anyway I needed to put the picture of the circles.)
So, now what I did first was basic stuff and found that the first angle of refraction was $30^\circ$.
After that, some math showed that the line was tangent to the inner sphere.
After this I can actually resort to lengthy non-physics related, completely mathematical stuff wherein I will need to take help of equation of tangents and stuff but that would take up a lot of time and is not possible in the exam hall.
This is one of the best(intriguing) questions I have seen so far, in ray optics.
Any help on how to solve it (non-mathematically)?
 A: In real life the ray will hardly enter the sphere because most of it will be reflected. In this case the answer would be trivially 0°. But I suppose the authors still meant the ray to be traced through the spheres, so I proceed with this assumption.
This problem is actually quite possible to solve without special physical insight. The numbers given here are "nice" enough that it'd be unwise not to use this. So, you just need Snell's law and a bit of trigonometry.
Here's a sketch of the solution:
1 . Use Snell's law to find out that the angle of refraction. Notice that it's 30°, and the ratio of the outer and the inner radii is 2. As this ratio is the sine of the angle of refraction, the ray is tangent to the inner sphere. Draw the lines in the figure to see this.
2 . Do the same as in step 1, but for the inner sphere. You'll find the angle of refraction is once again 30°, so the ray is tangent to an auxiliary circle of radius $R/2$.
3a . Use symmetry with respect to the line that crosses the tangent point found in step 2 and the origin. You'll find that the ray exits the inner sphere becoming tangent to it, then exits the outer sphere becoming tangent to it. This symmetry argument is justified by Helmholtz reciprocity principle.
3b . Alternatively, if symmetry doesn't convince you enough, continue the ray to the point of exit from the inner sphere and prove that its incidence angle is 30°. Proceed accordingly.
4 . Carefully calculate the accumulated angles of deviation, match to the answers.
The diagram will look something like this (red lines are auxiliary, green is the ray, blue denotes the angles):

