I presume you mean "the radius of the circle formed by the surface of the water at different heights".
And, by the way, the problem is not "well formed". If you just punch a hole in the bottom, the tank will empty without any work being done. I presume that the problem says to empty the tank by pumping it out the top.
Here's how I would think about it:
Of course, the work done lifting a mass m a height h is mgh. Imagine a "layer of water" of thickness dz at height z. That layer will be a circle of radius, say, r. It will have volume \pi r^2 dz and so weight 98\pi r^2dz. That water has to be lifted a distance 6-z to the top of the tank: the work done on that one "layer" of water is \pi (6-z)r^2 dz.
Now we need to determine r2: x2+ y2+ z2= 36. Since r is measured parallel to the xy-plane, r2= x2+ y2= 36- z2. Putting that into the formula above:
\pi (6-z)(36-z^2)dz.
We need to integrate that from z= 0 to z= 6 to cover the entire tank:
\int_0^6\pi(6-z)(36-z^2)dz= \pi\int_0^6(216- 36z-6z^2+ z^3)dz.
That is not what you have. Did you forget the height lifted?