Acircular swimming pool has a diameter of 24 ft, the sides are 5 ft high, and the depth of the water is 4 ft. how much work is required to pump all of the water out over the side? (use the fact that water weighs .) 62.5 lb

The pumping work has the equation written below:
W = mΔh,
where m is the mass
Δh is the difference in height

We determine the mass from the density of water which is 62.5 lb/ft³.
Density = Mass/Volume
Mass = m = Density*Volume
m = (62.5 lb/ft³)(π/4)(24 ft)² = 9,000π
The lb is in lbm or pound-mass. To convert it to pound-force, lbf, we use the g/gc conversion factor which is equal to 1 lbf/lbm. Thus,

W = (9,000π lbm)(5 ft - 4 ft)(1 lbf/lbm) = 28,274 lbf·ft

Ithink the answer is c but then again i'm probably wrong