How many cubes have only two coloured faces?

In any plane, leave 4 sides cube and select (3 x 3 x 3) inter section. But
the cubes 2 x 2 x 1 give 2 less cube because that part we are already removed.
No. of cubes = (3 x 3 x 3) - 2 = 25.

Total no. of cubes = 5³ = 125,
Some cubes from different corners are removed and the number removed cubes are 2, 3, 4 and 4.
Remaining number of small cubes:
= 125 - 2 - 3 - 4 - 4 = 125 - 13 = 112

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
In this case we have to use red and green twice and same colour should be on opposite faces then required cube is given by 4 edges (but not corner), maximum number of cubes one edge is 6 - 2 = 4 so required number of cubes is 4 x 4 = 16

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
Maximum number of cuboid with red colour is possible when cube is painted with red colour in 3 sides with minimum number of common edges (which is equal to 2)
Hence required maximum value is 6 (5 + 5 + 4 - 2) = 72
For minimum number of such cuboid Red colour is used only once and minimum number of cubes in that case is 20
Hence required ratio is 72 : 20 = 18 : 5

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
In this case when k is maximum, one particular colour is used on there faces such that any two faces are adjacent to each other. Required number of cuboids will come from edges but not from vertex = 3 + 4 + 5 + 1 = 13

Each has Red faces on top layer = all edges cube = 2 + 2 + 2 + 2 = 8

Number of cubes with 3 face coloured = 4 (Bottom cubes) + 8 top cubes + 4 (column cubes) = 16

Initial total number of cubes = 343,
Number of cubes removed = 27
Smaller 27 cubes painted blue
Exposed faces of original big cube (3 faces with 9 cube on each face i.e total 27 cubes) painted with black
Since 7 corner (Vertices) of bigger cube is untouched hence they are painted with three faces.
Now consider the corner from where we have removed 3 x 3 x 3 cubes,
After removed 3 new corners of the bigger cube will be generated that will be painted with 3 faces and 8 corners from smaller cube of 3 x 3 x 3 painted with 3 faces.
So the such total number of such cubes is 7 + 3 + 8 = 18.

Initial total number of cubes = 343,
Number of cubes removed = 27
Smaller 27 cubes painted blue
Exposed faces of original big cube (3 faces with 9 cube on each face i.e total 27 cubes) painted with black
In original big cube number of faces with one colour is 3(6 -2)² = 48 (here we have considered only 3 untouched of big cube)
But here we have removed a cubes of the form of 3 x 3 x 3 and again put it back so out of three new exposed faces of big cube we will have 4 cubes in each face that is painted with one colour hence number of cubes from these three surfaces is 3 x 4 = 12
Now consider out of 3 x 3 x 3 cubes we will have 6 cubes (one in each face ) which are painted only one face.
Hence total number of cubes = 48 + 12 + 6 = 66

Initial total number of cubes = 343,
Number of cubes removed = 27
Smaller 27 cubes painted blue
Exposed faces of original big cube (3 faces with 9 cube on each face i.e total 27 cubes) painted with black
Without any changes number of cubes with no face colour is given by (6 - 2)³ = 64
Now because of removal of 3 x 3 x 3 cubes from one of the corner from each face that were not painted earlier got exposed and will get painted, so from 3 x 3 x 3 cubes 4 x 3 = 12 cubes got painted, and a similar number from 3 exposed faces of big cube got painted.
Total number of cubes with no face painted is 64 - 12 - 12 = 40