Rotating Star physics question

The mass of a star is 1.47x10^31 kg and it performs one rotation in 30.7 days. Find its new period (in /days) if the diameter suddenly shrinks to 0.63 times its present size. Assume a uniform mass distribution before and after.

For a sphere, radius of gyration k, is given by k^2 = 3R^2 /5
Apply conservation of momentum:
Momentum = Iω
1rev in 30.7 days = 2 π / 30.7 x 24 x 3600 = 2.369 x 10^ -6 rad/s = ω
Initial momentum = (1.47 x 10^31 x 3 x R^2 / 5) x 2.369 x 10^ -6 = 2.089 x 10^25 x R^2

New radius = 0.63R
New k^2 = 3 (0.63R)^2 / 5 = 0.2382R^2
New Momentum = I ω =1.47 x 10^31 x 0.238R^2 x ω= 3.499 x10^30 x R^2 x ω
Initial momentum = Final momentum
2.089 x 10^25 R^2 = 3.499 x10^30 xR^2 x ω(final)

ω (final) = 2.089 x 10^25 R^2 / 3.499 x10^30 xR^2
= 5.97 x 10^-6 rad /sec
= 2.369 x 30.7 / 5.97
= 12.29 days period

Angular momentum is conserved so when the radius of the star is reduced it will, "spin-up".

Conservation of angular momentum says I1 x w1 = I2 x w2
I of a uniform mass distribution sphere = 2/5 ma^2

2/5 m x a^2 x (2 x pi/T1) = 2/5 m x (0.63a)^2 x (2 x pi/T2)
a^2/T1 = (0.63a)^2/T2
T2 = T1 x 0.63^2 x a^2/a^2
T2 = T1 x 0.63^2
T2 = 30.7 x 0.3969
T2 = 12.18 days
T2 = 12.2 days

I'm not sure that the mass is needed for this as the mass is conserved.