For any rectangle with length $l$ and breadth $b$, the radius of it's circumcribed circle is $\frac{\sqrt{l^2+b^2}}{2}$. (Join two diagonal vertices to the center to see this) So, the required radius is $\frac{\sqrt{AB^2+BD^2}}{2}$. Observe that in $\Delta BDC$, $BD=AB\cos 30^{\circ}$. Again, inradius of $\Delta ABC$ is $\frac{AB}{2\sqrt{3}}=1$. Plugging all the values, the answer is $\frac{\sqrt{21}}2$.

@above, a rectangle with side length $l$ and width $w$ has a circumradius of $$\frac{\sqrt{l^2+w^2}}{2}$$