Answer

The center of gravity of the third block is given by:

cg3 := (2*(cg2 + 1/6) - 2/6) / 3;

The combined center of gravity of the top two blocks has been moved one-sixth of a unit to the right. The center of gravity of the bottom block is two-sixths of a unit to the left of the edge of the table. We multiply the center of gravity of the top two blocks by two because they are twice as massive as the bottom block, and we divide the sum by three (instead of two) because there are three blocks.

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