$8\times8$

Let's look at the black polygon's perimeter. In the beginning it's perimeter is $4$ .

• If we color a cell which has $1$ adjacent black square, then the perimeter grows by $2$ .

• If we color a cell which has $3$ adjacent black squares, then the perimeter decreases by $2$ .

So suppose yes, it could be. Then we have to color $63$ squares. Since by two coloring the perimeter grows by $4, 0$ or $-4$ , after the $62$ . coloring the perimeter will be divisible by $4$ , so after $63$ coloring, the perimeter won't be divisible by $4$ . (It will make $2$ remainder.) But the whole square's perimeter is $4\times 8=32$ , which is not possible.

So the answer is: No, it can't happen.

It is supposed that by yellow you mean white, so if it would be possible, then you must be able to place the last black square but since all others are black already, it is adjecent to 0 white squares: contradiction.