Answer
$I = 0.019~kg~m^2$
Work Step by Step
We can find the rotational inertia:
$I = \frac{1}{12}~M~L_1^2+M~(\frac{L_2}{2})^2+\frac{1}{12}M~L_2^2$
$I = (\frac{1}{12})~(0.20~kg)~(0.40~m)^2+(0.20~kg)(\frac{0.50~m}{2})^2+(\frac{1}{12})(0.20~kg)~(0.50~m)^2$
$I = 0.019~kg~m^2$
