Let F be a point in BCDE such that DEF is an equilateral triangle. Then, observe that AFEB is a parallelogram since AB is equal and parallel to EF, thus $AF = BE = ED = FE = FD$ , so $F$ is equidistant from points $A, E, D$ . Hence, $F$ is the center of the circle, which means that the radius is 2010.

Note: The above solution relies on "make this wonderful observation", which I typically do not like.

The proper way to find the circumcenter of $AED$ would be to take the perpendicular bisectors. However, it is not immediately obvious how/where the perpendicular bisectors of $AE$ and $AD$ would intersect. Hence, I present the above solution, instead of a tedious coordinate geometry solution.

To generalize

Let 2010=a

Angles ABE AND ACD are 150 degrees Applying cosine rule any one of triangles ABE or ACD we get length of AE & AD. Now by using abc/4R = area of triangle for triangle AED where R is radius of circumcircle and a,b and c are its sides. We get R = a