Answer
We can rank the disks according to the torque about the disk center:
$1 = 2 \gt 3$
Work Step by Step
We can write a general expression for the torque about the center:
$\tau = r\times F = r~F~sin~\theta$
$r$ is the displacement vector from the center to the point where the force is applied
$F$ is the force vector
$\theta$ is the angle between these two vectors
Note that the angle between the $r$ vector and the $F$ vector is $90^{\circ}$ in each case.
On Disk 1 and Disk 2, the force $F$ is applied on the outer edge, while on Disk 3 the force $F$ is applied on the inner edge.
We can rank the disks according to the torque about the disk center:
$1 = 2 \gt 3$
