You can solve it!!

based from the problem the diagonal of the square will become the legs of the right triangle that will be formed. The hypotenuse is equal to the radius of the circle.

$\angle CED=150^{\circ}$ hence major arc $CD=300^{\circ}$ and $\angle COD=60^{\circ}$

That makes $\triangle OCD$ equilateral hence radius is $1$

We can place the problem in the coordinate plane, with $A=(0,0), B=(1,0), C=(0,1), D=(1,1).$ .

$E$ forms and equilateral triangle with $A$ and $B$ , so $E= (\frac{1}{2},\pm\frac{\sqrt(3)}{2})$ . Interestingly, these cases give the same result, and I will only give the example of $E= (\frac{1}{2},\frac{\sqrt3}{2})$ .

We now have $C=(0,1), D=(1,1), E=(\frac{1}{2},\frac{\sqrt3}{2})$ are on the same circle. The center of the circle must be equidistant from $C$ and $D$ so the x-coordinate of the center must be $\frac{1}{2}$ .

Now, using the distance formula, we set the distance from the center to C equal to the distance to E:

$\frac{1}{2}^2+(y-1)^2=(y-\frac{\sqrt3}{2})^2$

$\frac{1}{4}+y^2-2y+1=y^2-\sqrt3+\frac{3}{4}$

$(2-\sqrt3)y=\frac{1}{2}$

$y=\frac{1}{2} \frac{1}{2-\sqrt3}=\frac{2+\sqrt3}{2}$

The distance from the center to E is just $\frac{2+\sqrt3}{2}-\frac{\sqrt3}{2}=\boxed{1}$