The polyhedron you create by that process is kind of hard to visualize. But if you look at just one face of the cube, you have a square pyramid.

The surface area of the pyramid (not including the base) is 4 times the area of one triangle. The area of a triangle is , where b (the base) is just a, and you can calculate h from the fact that the angle at the midpoint of a side is 45 degrees. The midpoint of a side, the center of the square base, and the apex of the pyramid form a 45-45-90 triangle, so you can calculate h.

So you have the area of 1 triangle times 4 to give the surface area of the pyramid (minus its base), times 6 faces on a cube gives you the surface area of the whole polyhedron.

If you can imagine 4 rhombi meeting at a vertex pointing straight up, there are 4 points where 2 rhombi meet and 4 points where there is only 1 rhombus. Now connect 4 more rhombi, oriented vertically, with the "upper" level of vertices now finished with 3 rhombi and the "middle" level also having 3 rhombi, but the smaller angles. Now make another group of 4 rhombi meeting at a vertex pointing straight down, and it connects so that the "middle level" is complete with 4 rhombi meeting at a vertex, and the vertically-oriented rhombi fit into the "lower level", with 3 rhombi per vertex.

Post again if you are still having trouble.