To solve this problem we will use Snell's Optical law which warns that the ratio of the sine of the angle of incidence to the angle of refraction is equivalent to the index of refraction of the material, therefore,
[tex]n = \frac{sin\theta_1}{sin\theta_2}[/tex]
[tex]n= \frac{sin45}{sin 12}[/tex]
[tex]n = 3.4[/tex]
Therefore the refractive index of Unobtanium is 3.4.
Note: I used the full breast values for 45 and 12 (approximately 10 significant figures)
