The moment of inertia of a uniform circular disc rotating about an axis perpendicular to the plane of the disc and passing through points O, A, B and C are $I_o$, $I_A$, $I_B$ and $I_C$ respectively, then

By the parallel axis theorem, the moment of inertia of the disc about an axis perpendicular to its plane and passing through a point at a distance $d$ from the center is given by:

$I = I_{cm} + Md^2$

Here, the center of mass is at O, so $I_{cm} = I_O$.

From the given figure, the distances of points O, A, B, and C from the center O follow the order:

$d_O < d_A < d_B < d_C$

Since the moment of inertia $I$ increases as the distance $d$ from the center of mass increases, the moments of inertia will follow the same order:

$I_O < I_A < I_B < I_C$

Answer: $I_o < I_A < I_B < I_C$