What I know about $\ce{^{.}CR3}$ radical is that it has both $\ce{sp^2}$ and $\ce{sp^3}$ character but the $\ce{sp^2}$ character dominates and so the radical is $\ce{sp^2}$ in nature.
So, suppose we have the above situation. In this, a radical is formed at the carbon of the 6 membered ring at which one $\ce{-CH3}$ is attached since the bromination is selective, but will it form enantiomers? That is, can the $\ce{^{.}Br}$ radical attack from both sides?
In this question, the answer given is (c), but I think that enantiomers can not form, as the electron of the radical would be present only on one side of the radical carbon, and so cannot migrate to other side.