imc

By placing the $2\times 3$ rectangle adjacent to the $3\times 4$ rectangle with the 3 side of the $2\times 3$ rectangle next to the 4 side of the $3\times 4$ rectangle, we get a figure that can be completely enclosed in a square with a side length of 5. The area of this square is $5^2=25$. Since placing the two rectangles inside a $4\times 4$ square must result in overlap, the smallest possible area of the square is $25$. So the answer is $\boxed{25}$.