jmc

Let $s$ be the length of a side of the square. Consider an isosceles right triangle with vertices at the centers of the circle of radius 2 and two of the circles of radius 1. This triangle has legs of length 3, so its hypotenuse has length $3\sqrt{2}$.



The length of a side of the square is 2 more than the length of this hypotenuse, so $s=2+3\sqrt{2}$. Hence the area of the square is $$s^2=(2+3\sqrt{2}{)}^2=\boxed{22+12\sqrt{2}}.$$