Answer
$I = \frac{1}{3}M(a^2+b^2)$
Work Step by Step
For a rectangular sheet rotating around the center, the moment of inertia is $I_{com} = \frac{1}{12}M(a^2+b^2)$.
Let $d$ be the distance from the center of the sheet to one corner.
$d^2 = (\frac{a}{2})^2+ (\frac{b}{2})^2$
We can use the parallel axis theorem to find the moment of inertia when the axis is at one corner.
$I = I_{com}+Md^2$
$I = \frac{1}{12}M(a^2+b^2)+ M[(\frac{a}{2})^2+ (\frac{b}{2})^2]$
$I = \frac{1}{12}M(a^2+b^2)+\frac{1}{4}M(a^2+b^2)$
$I = \frac{1}{3}M(a^2+b^2)$
