3-D Tiling

Let A, B, C, D, E and F be the tiles and x be the filled center cube. Here we go:

Every $2 \times 2 \times 1$ tile will occupy one side of the center cube. Since the center cube has 6 sides, exactly 6 tiles will fit. If each of these tiles keeps the tile in the same corner, then we know that these tiles will not overlap.

The volume of the 3 x 3 x 3 cube, as stated, is 26 cube units since the center cube is occupied. The volume of the 2 x 2 x 1 cube would be 4 cube units. So, if you divide 26 by 4, the quotient would be equal to 6, with a remainder of 2 units. So, six 2 x 2 x 1 cubes would fit in the region without overlapping.

This is just an idea. I think this is the solution because this is the first of infinitely many configurations of $2\times2\times1$ tiles that would give us a cube with odd dimensions (not necesarilly filled, but what is important here is that it gives the dimmensions). The key is parity. When we try to construct a cube (different from the easy $2\times2$ cube), we first Take two tiles and connect them forming an "L". Then notice that the dimensions of this figure are $2\times2\times3$ . Then connect another "L" but in opposite direction and you form a $3\times3\times3$ , figure. However, another "L" can be incorpored to the figure keeping the same dimmensions, leaving three spaces that form a diagonal through the cube. I know this maybe isn't that clear, but at least I tried... Maybe someone can state what I'm traying to say more clearer. Sorry for my bad english :)

[This was a wrong solution.]