Answer
$\omega=0.075rev/day$
Work Step by Step
Angular momentum is conserved, so $$I_0\omega_0=I_f\omega_f$$
Before the explosion, the star is a solid sphere of radius R, so $I_0=\frac{2}{5}MR^2$
After the explosion, the star is a thin-walled shell of radius 4R, so $I_f=\frac{2}{3}M(4R)^2=\frac{32}{3}MR^2$
Therefore, $$\frac{\omega_f}{\omega_0}=\frac{I_0}{I_f}=\frac{3}{80}$$ $$\omega_f=\frac{3}{80}(2rev/day)=0.075rev/day$$
