Answer
$I = 2MR^2$
Work Step by Step
$I_{com} = MR^2$
We can use the parallel axis theorem to find the moment of inertia when the axis of rotation is on a point on the edge of the hoop. Note that $d$ is the distance from the center of mass to the axis of rotation.
$I = I_{com}+Md^2$
$I = MR^2+MR^2$
$I = 2MR^2$
