That's not how I see it. (I believe you have the same interpretation that I suggested, that the axle of the disk is fixed in the lab frame.)
Let me take it a bit further. Let ##\vec d## be the position vector of the axle relative to the platform's axis, ##\vec r## the position vector of some disk element relative to the disk's centre, and ##\vec s ## be the position vector of the same element relative to the platform's axis. So ##\vec s= \vec d+\vec r##.
If ##\vec \omega## and ##\vec \omega'## are the rotation vectors of the platform and disk respectively then at the element the velocities are ##\vec s\times\vec \omega## and ##\vec r\times\vec\omega'##. The relative velocity is ##\vec v=\vec r\times\vec\omega'-\vec s\times\vec \omega##, and the unit vector in that direction is ##\frac{\vec v}{|\vec v|}##. Multiplying that by a constant and taking the cross product with ##\vec r## produces the torque on the disk per unit area of the element due to that element. Finally, multiplying that by the area of the element and integrating over the disk gives the overall torque. The task is to find the value of ω' that makes that zero.
So that has taken the analysis somewhat further, yet still the hard part, the integral, lies ahead.
@jbriggs444 found the identical problem online as an Olympiad problem. It sketches the solution so vaguely that I am unconvinced their solution is valid.
Edit: wrt the proposed answer of -ω (the answer is not given at the olympiad link, so maybe that should not be trusted), it is fairly clear that for d=0 it would be +ω, as d tends to infinity it would tend to 0, and that for every other d it would be between those values.