Answer
$8.4\times10^3 kg \frac{m^2}{s}$.
Work Step by Step
First find the moment of inertia of the system, which is the sum of the individual moments.
$$I=\frac{1}{2}(110kg)(4m)^2+(50kg)(4m)^2$$
$$I=1680kg\cdot m^2$$
Now find the angular momentum.
$$L=I \omega=(1680kg\cdot m^2)\frac{0.80rev}{s}\frac{2 \pi rad}{rev}=8.4\times10^3 kg \frac{m^2}{s}$$
