Volume of sphere, V=(4/3)πr³ so r³=3V/(4π) and r=∛(3V/(4π)) where r=radius. From this, r=9.1983m approx given V=3260 cubic metres.
If h=depth of water then r-h is the distance between the centre of the sphere and the surface of the water.
So, by Pythagoras, a, the radius of the water’s circular surface=√(r²-(r-h)²)=√(h(2r-h)).
Put r=9.1983 and h=5.25, a=8.3078m approx.
I’m puzzled about this because the water level should be about 4.98m not 5.25m. It wasn’t necessary to specify both the height and volume of water (?) Am I missing something?