Only the 8 corner cubelets of the original large cube will have three painted faces.
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Draw a three dimensional cube, showing three faces. Divide each face into 6 x 6 =36 squares, representing the faces of the little cubes. Then by inspection it is obvious that no little cube can have more then three painted faces and they will be at the eight corners. Only the edge cubes (not counting the corners) can have two painted faces and there are 48 of then. Finally, only the inner cubes on the surfaces will have one painted face and there are 96 of them. The other 64 cubes are in the inner core and will have no painted surfaces. SO: 6 = 0 ; 5 = 0; 4 =0 ; 3 = 8 ; 2 = 48 ; 1 = 96 . uuuuuu.....
Answer = 8. The only small cubes with paint on three faces are those that occupied the corners of the original cube.
This is kind of neat. I drew a three dimensional cube. Then divided the faces into 16 squares per face. Then started counting which of those little cubes would have a painted face. I get: 1 red face - 24 cubes 2 red faces - 24 cubes 3 red faces - 8 cubes 4, 5, & 6 red faces - 0 cubes
It depends on how the rectangular solid is built. If the side of a cube is "a", then your original solid could be a x a x 54a, or a x 2a x 27a, or 2a x 3a x 9a or one of quite a few others, and the answer depends on the one you specify.
There would be 8 will 3 faces painted red, 24 with 2 faces painted red, 24 with 1 face painted red, and 8 that have no faces painted red.