Twenty-seven unit cubes are each coloured completely black or completely red. The unit cubes are assembled into a larger cube. If $\frac{1}{3}$ of the surface area of the larger cube is red, what is the smallest number of unit cubes that could have been coloured red?
Since $\sqrt[3]{27}=3$, the dimensions of the larger cube must be $3 \times 3 \times 3$.

Therefore, each side of the larger cube has area $3 \times 3=9$.

A cube has 6 faces, so the total surface of the cube is made up of $9 \times 6=54$ of the $1 \times 1$ squares from the faces of the unit cubes.

Since $\frac{1}{3}$ of the surface area is red, this means $\frac{54}{3}=18$ of these unit squares must be red.

The unit cube at the centre of the larger cube has none of its faces showing, the 6 unit cubes in the centres of the outer faces have exactly 1 face showing, the 12 unit cubes on an edge but not at a corner have 2 faces showing, one of each of two adjacent sides, and the 8 unit cubes at the corners each have 3 faces showing.

For any unit cube, there are either $0,1,2$, or 3 of its faces showing on the surface of the larger cube.

This means at most three faces of any unit cube are on the surface of the larger cube. Thus, there must be at least 6 cubes painted red in order to have 18 red $1 \times 1$ squares on the surface of the larger cube.

There are 8 unit cubes on the corners, so if we colour exactly 6 unit cubes red and the other 21 black, then arrange the cubes into a $3 \times 3 \times 3$ cube so that the 6 red unit cubes are at corners, there will be exactly 18 of the $1 \times 1$ squares on the surface coloured red.

Therefore, the answer is 6 .

ANSWER: 6