Answer
$I = 1.8~kg~m^2$
Work Step by Step
We can use the parallel axis theorem to find the rotational inertia of the hoop:
$I = mR^2+mR^2 = 2mR^2$
We can find the rotational inertia of the square:
$I = 0+mR^2+\frac{1}{3}mR^2+\frac{1}{3}mR^2 = \frac{5}{3}mR^2$
We can find the rotational inertia of the structure:
$I = 2mR^2+\frac{5}{3}mR^2$
$I =\frac{11}{3}mR^2$
$I =\frac{11}{3}(2.0~kg)(0.50~m)^2$
$I = 1.8~kg~m^2$
