#### Answer

The angular momentum is $2.1~kg~m^2/s$

#### Work Step by Step

We can express the angular velocity in units of rad/s;
$\omega = (120~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$
$\omega = 4\pi~rad/s$
We can find the moment of inertia of the bar as;
$I = \frac{1}{12}MR^2$
$I = \frac{1}{12}(0.50~kg)(2.0~m)^2$
$I = 0.167~kg~m^2$
We can find the angular momentum about the axle:
$L = I~\omega$
$L = (0.167~kg~m^2)(4\pi~rad/s)$
$L = 2.1~kg~m^2/s$
The angular momentum is $2.1~kg~m^2/s$.