Divide it into $25$ squares of side $2$, and divide each square into four pieces by drawing two perpendicular lines through its center. Each quadrilateral has two opposite $90^{\circ}$ angles, so they are cyclic. If the lines are parallel to the sides of the squares, the quadrilaterals will have circumradii $\sqrt{2}$, and in the limit case where the lines approach the diagonals of the square, they will have circumradii $2$. Thus, in some intermediate case, the quadrilaterals will have circumradii $\sqrt{3}$, as desired. [asy][asy] draw((-1, -1)--(-1, 1)--(1, 1)--(1, -1)--cycle); draw((sqrt(2)/2, -1)--(-sqrt(2)/2, 1)); draw((1, sqrt(2)/2)--(-1, -sqrt(2)/2)); draw((-1, -sqrt(2)/2)--(sqrt(2)/2, -1), dashed); label("$\sqrt{3}$", ((-1 + sqrt(2)/2)/2, (-1 - sqrt(2)/2)/2), dir(90)); [/asy][/asy]