Answer
Choice D, $\frac{T}{\sqrt{3}}$.
Work Step by Step
We see from equation 13.12 how the period T depends on the mass of the central body (i.e, here, the mass of the star): $T=\frac{2\pi r^{3/2}}{\sqrt{GM}}$.
Increasing the star’s mass, M, by 3 will produce a new factor of $\sqrt{3}$ in the denominator, and reduce the period by that factor.
