It is easy to know that the axial section of a cone is a regular triangle, and the radius of the ball is the radius of the inscribed circle of the regular triangle of the axial section.

If the side length of a regular triangle is 2R, the height of the cone is h=√3R, and the radius of the sphere is r = (1/3) h = (√ 3/3) r.

The volume of the cone v cone =( 1/3)πR? h=(√3/3)πR？

The volume of the ball v ball =(4/3)πr? =(4√3/27)πR？

V remainder =V cone -V sphere =(5√3/27)πR?

So V- residual /V- cone =5/9.

That is, the remaining water in the container is 5/9 of the original.