Question 58209
I believe this is a higher thinking question, where you have to do a step removal process.

knowing a cube each side has 12 rows of 12 squares, when you subtract the outside squares (they have paint on 2 and 3 sides, you get 10 rows of 10 squares

by 6 sides there are 600 squares with one side painted

each corner of the square would have 3 sides painted, as there are 8 corners, 8 cubes have 3 sides painted.

now all we have left are the outside painted cubes, because the corners are missing, there is 10 per row, (12 per row subtracting 2 corners in a row=10 per row) 

there are 12 rows on a cube, (4 on the top, 4 on the bottom, and 4 on the sides, because they all share sides we can not state 4X6 sides we have to logically subtract the sides.

because there are 12 rows with 10 in a row, there is 120 cubes with 2 sides painted.

last, there are 10 rows, 10 columns, and 10 stacks left, hence there is 1000 unpainted cubes. here is the completed list.

8=cubes with 3 painted sides
120=cubes with 2 painted sides
600=cubes with 1 painted side
1000=cubes with no painted sides.
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1728 cubes in total

check, 12x12x12=1728, so we have accounted for all the squares.