Answer
$8.7\space Mm$
Work Step by Step
Let's assume this star is a solid sphere. In this case, there's no external torque. So its angular momentum is conserved.
We can write,
$I_{1}\omega_{1}=I_{2}\omega_{2}-(1)$
For a solid sphere, Rotational Inertia $(I)=\frac{2}{5}MR^{2}-(2)$
$(2)=\gt(1)$
$\frac{2}{5}MR_{1}^{2}\omega_{1}=\frac{2}{5}MR_{2}^{2}\omega_{2}$
$R_{2}=R_{1}\sqrt {\frac{\omega_{1}}{\omega_{2}}}$ ; Let's plug known values into this equation.
$R_{2}=7.1\times10^{3}m\sqrt {\frac{21.9\space rpm}{\frac{rpm}{49.3\times24\times60}}}=8669.1\times10^{3}m$
$R_{2}=8.7\space Mm$
