The steel ball will continue to float on the mercury, but the ball will float slightly higher because of buoyancy due to the mass of water it displaces.
Doing the sums for the cases before the addition of the water and for the case where the water is deep enough to completely cover the ball:
(These are based on the specific gravity of mercury, steel and water being 13.56, 7.85, and 1 respectively)
Before water addition: roughly 7.85/13.56 (57.9%) of the volume of the ball will be below the surface of the mercury
After water addition that entirely covers the ball: roughly 6.85/12.56 (54.5%) of the volume of the ball will be below the surface of the mercury
The above calculation is based on the idea that both the mercury and the ball are displacing water, so their water-buoyancy corrected specific gravities are reduced by 1.
Given that this may not be intuitive to all readers, her is a checksum to demonstrate validity:
Mercury mass displaced = 6.85/12.56*13.56/7.85 of the mass of the ball
Water mass displaced = (1 - 6.85/12.56)/7.85 of the mass of the ball
Total mass displaced by the ball = (7.39582 + 0.454618) = 1 x the mass of the ball