Me with this problem with steps.

I found this on Yahoo answers 18 different sets.

There are 6 faces, so that's 6 sets. 

If we draw two parallel diagonals on two opposite faces, their endpoints are another set of four coplanar points. There are 3 pairs of opposite faces, and each has two such sets of diagonals, giving us another 6 sets. 

Now consider the four diagonals of the cube. They all meet at a single point in the center of the cube. Therefore, any pair of them represent a pair of intersecting lines and so define a plane. So any pair of these diagonals produces yet another set of four coplanar vertices. 

There are four such diagonals, and there are 
4C2 = 4! / (2! 2!) = 4 * 3 / 2 = 6 different combinations comprised of two of them, so that's another 6. 

That gives us a total of 18 different such sets of four coplanar vertices of the cube. No other combinations of four vertices are coplanar.