We will start with counting polyominos of smallest area. There is only 1 polyomino of area 1, 1 polyomino of area 2, and 2 polyominos of area 3. Thus, the smallest area of a rectangle that consists of 8 polyominos is $1 \cdot 1 + 1 \cdot 2 + 2 \cdot 3 + 4 \cdot 4 = 25$. But then it has to be of a size $25 \times 1$ (since $5 \times 5$ square doesn't satisfy the condition). It is clear for such a rectangle that it can only be split into rectangles of size $k \times 1$ and, but such a rectangle has an area at least $1+2+...+8=36 > 25$. Thus, the smallest possible area is 26. An example of splitting a $13 \times 2$ rectangle is shown on Fig. 18.