What would be the approximate force required on the perimeter of a 30 ft diameter table that weighs 200,000 lbs to move it? The table is sitting on twelve wheels (10" diameter each) that are mounted to ball bearings.
The rolling resistance coefficient (RRC) for rubber automobile tires is about 0.01 (less if the tires are inflated more). Thus for a car weighing 3000 pounds, the horizontal force needed to push the car (overcome the RRC) is about 30 pounds.
For your case, I would guess about 0.01 x 200,000 pounds = 2,000 pounds tangential force on the perimeter of the table.
I don't think you are going to support 200,000 lbs on 12 10-inch-diameter rubber tires. The weight on each tire would be about 7.5 tons. Compare "one 10-inch diameter wheel" with the wheels and tires on a typical 7-ton road truck!
The question is impossible to answer unless we know more about the wheels, and what surface they are running on. You would probably need a specially built track to support that weight with such a small contact area. Otherwise, the most of the force will be overcoming the deformation of the ground and/or the wheels, not overcoming rolling resistance or friction in the bearings.
There are plenty of mobile engineering structures that weigh more than 100 tons. That is only the weight of three large "standard" road trucks. But supporting them on twelve 10-inch diameter wheels is not so common.
The gross weight limit of a standard US rail freight car is about 286,000 pounds, supported on eight 33" diameter steel wheels (35,750 pounds each). The rolling resistance coefficient of steel wheels on a steel rail (without deformation) is about 0.1%.