This looks like a "conservation of angular momentum" problem.
The angular momentum of the star (moment of inertia (I) times angular speed (w)) of the star is the same before and after the shrink happens. What changes is the size of the star and thus the distance that each part of the star is away from the center. In other words, the moment of inertia changes.
The moment of inertia of any object is some constant (k) times the objects mass, times some distance element:
I = k*m*r^2
The reason I'm being general here is that it turns out we don't even need to know the mass of the star nor that constant 'k' since we are already given the fraction to which the value of 'r' shrinks. In this case, r becomes .37 times what it was before the shrink. In the expression from moment of inertia (I) the r is squared, so if we shrink it to .37 of what it was, the moment of inertia is changed by the square of that.
I (after shrink) = I (before shrink) * 0.37^2 = I (before shrink) * .1369
So the moment of inertia of the star is .1369 times what it was before shrinking. Since angular momentum is conserved:
I (before) * w (before) = I (after) * w (after)
This means that:
w (after) = I (before) * w (before) / I (after)
We already have that the ratio I (before) / I (after) is 1/.1369 (the inverse of that ratio above) so the angular speed must be about 7.3 (1/.1369) times what it was before shrinking. We're being asked what the new period of rotation is, and we've found the amount that the angular speed changed. Period and angular speed (or angular frequency) are inverses of each other, so if the angular speed increases by a factor of 7.3 the period must shrink by a factor of 7.3. We're given that the period before the shrink is 21.9 days, so we just need to divide that by 7.3 (or multiply by 1/.1369).
Period after shrink = 21.9 days / 7.3 = 3 days
So the star only takes 3 days to rotate after shrinking. This is because it rotates more quickly after its radius (and diameter) decrease in accordance with conservation of angular momentum.
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