What is the 6th side?

Each side of the six faces of the cube coincides with another one, thus there must be an even number of every combination of green and red.

First, we notice that in the five given faces there are $3$ red sides (no green segment). Hence the sixth face must have such a side. This way we are left with three choices: A, B and E.

Secondly, we see that in the given faces there are $3$ sides of the form RRG i.e. $\frac{2}{3}$ of the side is red and the rest $\frac{1}{3}$ of the side is green. Thus, the sixth face must have such a side, so option B is discarded and we are left with options A and E.

The only difference between A and E is the colour of the central cube, which is green in A and red in E. in every vertex of the $3\times 3\times 3$ cube there are $3$ squares of the same colour that meet. In the given faces we find $5$ green small squares and a sixth is on both A and E. These squares correspond to $2$ green small cubes. Furthermore, we have $14$ middle small green squares in total either we choose A or E. These give $7$ more small green cubes. Finally, there is one central green square in the given faces which corresponds to one more green cube, summing to a total of $2+7+1=10$ green cubes. In order to reach the number $12$ for the green cubes we need the invisible cube in the center of the big cube to be green, as well as one more cube in the center of the sixth face. The latter verifies that option $\boxed{A}$ is the correct answer.

Consider the full $3\times 3\times 3$ cube. Any green cube located at a corner shows $3$ green faces. Any on an edge shows $2$ green faces; any in the middle of a face shows just one green face. There also may or may not be a green cube at the centre of the $3\times 3\times 3$ cube.

On the five given faces, we have $5$ green squares in a corner; $12$ on an edge; and $1$ in the centre of a face.

Say the missing face has $x$ green squares in a corner, $y$ on an edge, and $z$ in its centre. The total number of visible green cubes is then $T=\frac13 (5+x)+\frac12 (12+y)+1+z$

Since there are $12$ green cubes in total, and at most one can be hidden, $T$ must be $11$ or $12$ .

We can now look at which of the answer options might work:

Options with a non-integer number of green cubes can be dismissed immediately. Of the others, only option A fits all the criteria.

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Had we not had the answer options, this approach wouldn't be enough to solve the question; but it would still give some information about which parts of the missing face could be green.