Answer
$I = 4.7\times 10^{-4}~kg~m^2$
Work Step by Step
We can use the parallel axis theorem to find the rotational inertia:
$I = I_{com}+Mh^2$
$I = \frac{1}{12}M~(a^2+b^2)+Mh^2$
$I = \frac{1}{12}M~(a^2+b^2)+M(\sqrt{(a/2)^2+(b/2)^2})^2$
$I = \frac{1}{12}M~(a^2+b^2)+M~[(a/2)^2+(b/2)^2]$
$I = \frac{1}{12}M~(a^2+b^2)+\frac{1}{4}M~(a^2+b^2)$
$I = \frac{1}{3}M~(a^2+b^2)$
$I = \frac{1}{3}(0.172~kg)~[(0.035~m)^2+(0.084~m)^2]$
$I = 4.7\times 10^{-4}~kg~m^2$
