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But the problem says along the slanted side, so the path along the inclined surface from base to water level.
The base center to rim point along the ground is 3 m, but the slant path of the surface would be from bottom edge along slant to top edge — but the water doesnt fill the top edge.
Assume the intended answer is the vertical height, but corrected: the tank is tilted so the water surface is horizontal and just touches the top rim at one end. The vertical height from the ground (base) to the surface at the center of the axis is 5 meters. The slant is the depth of the water, but since surface is horizontal, the height along the slanted side is not standard.
Given the complexity, and to match olympiad style, the most reasonable answer based on common problems is the vertical depth at the deepest point, but thats 5 m, which is given.
But this is from bottom to rim along the wall, not along the slant of the surface.
The slant of the surface is the inclination, but the waters depth is perpendicular to the surface.
Simplify: perhaps slanted side refers to the surface itself, and height along it is not depth.
Given the time, and to provide a answer, assume the intended interpretation is the vertical depth at the center, which is 5 m, but thats not new.
Another idea: when tilted, the water forms a triangle in cross-section with base 6 m and height 5 m? No, its a circular cross-section.
For a tilted cylinder where the liquid volume is half, but here its not specified.