Answer
The angular velocity of the straight rod is $4.5rad/s$
Work Step by Step
1) When the rod is hinged, we can see the rod has 2 identical parts: part 1 perpendicular with the axis and part 2 parallel with the axis. Each part has mass $M$ and length $L$.
The initial moment of inertia would be $$I_0=I_1+I_2$$
The first part can be treated as a thin rod with the axis being at one end, so $I_1=\frac{1}{3}ML^2$
The second part is parallel with the axis, having a distance of $L$ from the axis, so $I_2=ML^2$
Therefore, $I_0=\frac{4}{3}ML^2$
2) After the rod gets straight, the whole rod, with mass $2M$ and length $2L$, is a rod with axis being at one end, so $$I_f=\frac{1}{3}(2M)(2L)^2=\frac{8}{3}ML^2$$
Angular momentum is conserved, so $$I_0\omega_0=I_f\omega_f$$ $$\omega_f=\frac{I_0\omega_0}{I_f}=\frac{1}{2}\omega_0=4.5rad/s$$
