So, as a radiator? Probably not too terrible.
I recall evaluating this, and the hot spot (effective active area) goes as the ratio of lateral conductivity to specific convection rate (assuming a linear W / (m^2 K) figure). The solution involves a Bessel function (assuming a circular heat source in the middle, and a circular boundary to the plate -- hence the radial symmetry, and the appearance of this function not being too surprising), but we don't need to evaluate it to get the aspect ratio.
The number is pretty reasonable, like for typical PCB (2oz pours) the effective radius is 5cm or so, so I think I had done that right. Though it's been long enough I don't recall all the setup or the exact formula anymore, just that it's reasonable in the way you'd expect...
So, substitute whatever the average conductivity of that PCB is, with your steel plate (material and thickness), and scale the radius accordingly.
Anyways, the K/W then follows from the effective area. Mind, the source (hotspot) temperature varies with its radius, ultimately going inverse with it (since, you put 100W into a pinpoint and it's going to turn to plasma, after all). So it does assume a big enough spot to start with. (Values are pretty reasonable for conventional size semiconductors, not much surprise there.)
60C is a pretty low ask though. A hundred watts maybe? An aluminum heatsink will definitely be more compact (and cheaper, if you didn't have the plates already that is), and a fan will help greatly. (Mind, the fan increases convection coefficient, thus shrinking the hotspot, and so too the active area -- dissipation still goes up quickly with airflow at first, as the boundary layer shrinks, but eventually the rate slows as the hotspot shrinks. I think.)
Tim