Answer
$I = 0.097~kg~m^2$
Work Step by Step
We can use the parallel axis theorem to find the rotational inertia:
$I = I_{com}+Mh^2$
$I = \frac{1}{12}ML^2+Mh^2$
$I = \frac{1}{12}(0.56~kg)(1.0~m)^2+(0.56~kg)(0.30~m)^2$
$I = 0.097~kg~m^2$
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