Answer
$I=\frac{3}{2}MR^2$
Work Step by Step
The parallel axis theorem states that $$I=I_{cm}+Mh^2$$
For a solid cylinder of radius $R$, we have $I_{cm}=\frac{1}{2}MR^2$
The axis in this exercise lies on the surface of the cylinder, so the distance between that axis and the axis passing through the center of mass $h=R$
$$I=\frac{1}{2}MR^2+MR^2=\frac{3}{2}MR^2$$
