The radius of the cone bottom water surface is 1* root 3= root 3. At this time, h== root 3* r = 3 (1/2) * r = root 3* root 3=3.

Then v1= π * r * r * h/3 = π * h * r 2/3 = π * 3 * 3/3 = 3 π.

Sphere volume: V2 = 4 π * R * R/3 = 4 * π * R 3/3 = 4 * π/3.

So the volume of water is: V=V 1-V2=3π-4π/3=5*π/3.

After the ball is taken out of the cone, assuming that the radius of the cone horizontal plane is r, then H= root 3 * r = 3 (1/2) * r.

Then v = π * r * r * h/3 = π * r * r * 3 (1/2) * r/3 = π * r 3/(3 (1/2)) = 5 * π/3.

The solution is: r = (5/(3 (1/2)) (1/3) h =15 under the cubic root sign, that is, 15 (1/3).