Volume of one ball=4π×53/3=500π/3 cc. Three such balls have a combined volume of 500π cc, which increases the overall volume of balls plus water. When the balls are stacked one on top of the other the total height of the stack cannot exceed 3 ball diameters=30cm so, since the height of the water in the vessel is 35cm, it accommodates the stack, so all the balls are surrounded by water, assuming they don't float.
The water level is at 35cm. The increase in water level is 500π/(36π)=13.89cm approx., where 36π is the area of the base of the vessel. So the water level rises to 35+13.89=48.89cm.
Another way to solve this is to calculate the volume of the water=1260π cc (πr2h, where h=35cm, r=6cm). Add to this the volume of the balls=1760π, then divide by the base area: 1760π/36π=48.89cm, the increased water level.