0 - you said it's only 1 cube even if it's a big one
it has six faces :)
It depends on how they are connected, i.e. how many are connected to each other. If they are connected in a line, with just one connection between any two cubes, the rectangle formed still has 6 faces, four of them rectangular (10 of the original 36 cubic faces are hidden between cubes, leaving 26 showing). If the maximum number of faces are connected, with 4 of the cubes as a larger cube and the other two attached together to it, there would be 3 large irregular faces (sides and back), 2 large cubic faces (bottom and front) and 3 rectangular faces (half cubes) as seen from the top. The total would be 8.
6
488 cubes
8
0 sides are not painted! So all the cubes have three faces painted!
Only the 8 corner cubelets of the original large cube will have three painted faces.
All the little cubes except the one at the middle will have 1, 2 or 3 faces with paint on them, ie 26 of them will have some paint on them: 0 faces painted - 1 cube 1 face painted - 6 cubes (the centre ones of each side) 2 faces painted - 12 cubes (the ones in the centre of where two sides meet) 3 faces painted - 8 cubes (the ones in the corners).
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.
I am assuming that your question states that the top and sides are painted red but the bottom isn't. If so, the answer is:- There are no cubes with 4 faces painted, the most that can be painted is 3 for the ones on the corners. There are 4 corners at the top of the cube that will have their top and 2 sides painted. Therefore there are 4 cubes with 3 painted. The cubes at the corners on the second and third row down will have 2 faces painted, as will the middle cubes on the top row so there are 12 of them in this puzzle. The cube in the middle of each painted face will have just one face painted so this is 5 (assuming the bottom isn't painted). The cubes in the middle of the bottom row will also have one face painted. This brings the total to 9 That is the total of cubes that have paint on them... 4+12+9 = 25 There are 27 cubes in your puzzle so only 2 have no painted faces. The cube right in the middle and the cube in the middle of the bottom layer.
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
54
Each cube has 6 faces. So 4 cubes together would have 24 faces.
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.....
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.
0 - you said it's only 1 cube even if it's a big one