The moment of inertia of a uniform circular disc about a diameter is $I$. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be 

Moment of inertia of disc about a diameter 

$=\frac{1}{4}M{R}^2=I$ (given)  
$\therefore M{R}^2=4I$
Now moment of inertia of disc about an axis perpendicular to its plane and passing through a point on its rim
$=\frac{3}{2}M{R}^2=\frac{3}{2}\left(4I\right)=6I$