Answer
$\omega = 6.06~rad/s$
Work Step by Step
We can find an expression for the rotational inertia:
$I = \frac{1}{3}ML^2 + ML^2 = \frac{4}{3}ML^2$
After the object falls from the horizontal position to the vertical position, the center of mass of the middle rod falls a distance of $\frac{L}{2}$ while the rod farthest from the axis falls a distance of $L$
We can use conservation of energy to find the angular speed:
$K = U_1-U_2$
$\frac{1}{2}~I~\omega^2 = Mg(\frac{L}{2})+MgL$
$(\frac{1}{2})~(\frac{4}{3}ML^2)~\omega^2 = \frac{3MgL}{2}$
$\omega^2 = \frac{9g}{4L}$
$\omega = \sqrt{\frac{9g}{4L}}$
$\omega = \sqrt{\frac{(9)(9.8~m/s^2)}{(4)(0.600~m)}}$
$\omega = 6.06~rad/s$
