Construct the square $ABCD$ with sidelength $6$ s.t. $C$ and $D$ lie outside the circle. Since $OC=OD$ by symmetry, there exists a homothety about $O$ that sends $C$ and $D$ to $C'$ and $D'$ s.t. $C'$ and $D'$ both lie on the circle. Square $A'B'C'D'$ will be inscribed in the sector. Using pythagorean theorem, we get that $OC=OD=\sqrt{10^2+3^2}=\sqrt{109}$. So the ratio of the areas of $ABCD$ to $A'B'C'D'$ is $\frac{109}{25}$ Since the area of $ABCD$ is $36$, the area of $A'B'C'D'$ is $\frac{900}{109}$