To check your answer, consider an arbitrarily small "cell" of water
inside the tank after it is filled.
If the water in that cell has mass $\mathrm d m$ and is at height $y$
above the ground, the amount of energy you have to expend to raise it from the
ground to its final position is $y\;\mathrm d m.$
For each such cell of water above the level of the center of the tank,
there is an equal-volume cell of water inside the tank that is the
mirror-image reflection of the first cell through the
horizontal plane in which the center of the tank lies.
And vice versa, for each cell below that plane there is a mirror image
of that "cell" in the tank above the plane.
The entire tank can be partitioned into pairs of tiny "cells"
reflected through that central plane.
Let $h$ be the height of the center of the tank above the ground, and
consider a pair of cells of which the upper cell is at height $h + \Delta y.$
The other cell is at height $h - \Delta y.$
The total potential energy you must invest in those two cells is
$$ (h + \Delta y)\,\mathrm d m + (h + \Delta y)\,\mathrm d m = 2h\;\mathrm d m.$$
That is, the energy invested in those two cells (of combined mass $2\,\mathrm d m$)
is the same as the energy required to lift a mass $2\,\mathrm d m$
to a height $h$ from the ground.
We can therefore simplify the calculation by replacing every such pair of cells with
one such mass, that is, we take the entire mass of water inside the sphere and
substitute an exactly equal mass at a height of exactly $h.$
Your answer will be correct if it equals $mh,$ where $m$ is the mass of water in the tank.
I suspect your instructor did not intend you to use this method, so I recommend to
either keep this as a mere check on the result you get by the intended methods,
or at most show it as an alternative method after deriving the result by those