jmc

Call the center of the circle O, and call the point where the radius of the circle bisects the chord E. Thus, the line segment from the center of the circle to point E has length 2, and we have $△ODE$ with a leg of 2 and a hypotenuse of 3. Thus, the other leg, DE has length $\sqrt{5}$, and since DE is $\frac{CD}{2}$, we have $CD=\boxed{2\sqrt{5}}$.