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.