Ahhh, the 30 - y comes into play because as you pump, the work required to pump it in changes (linearly)
Your trying to figure out the work required to pump from 0 to dy, then from dy to 2dy, then from 2dy to 3dy...
This makes sense because as you pump more kerosene into the tank it fills up and the contained kerosene feels a weight force and exerts that force as W on the kerosene you're pumping in, which requires you to do more work to get it in there.
It appears as though they are saying there is no work required to pump it in (without the weight of the kerosene in there. So basically you're integrating to find the work done on the "pump" by the kerosene in the container.
##MATH##
What's the weight force from an arbitrary amount of kerosene?
##\rho V = \rho Ay##
but, for any given amount of kerosene what is the work required to "lift" that kerosene to the top?
keeping in mind ##W = \int F (dot) dr##
we have ##W =\int \rho A dy (dot)## (distance from the height of the kerosene [y] to the top)
This happens to be a displacement parallel to the force, so we can drop the dot product and what we have is
##W= \rho A \int_{a}^b (30 - y)dy##
more specifically
##W = \rho A \int_{0}^{30}(30-y)dy = 5120\pi (30y-\frac{y^2}{2})|_0^{30}##
##=5120(450) \pi \approx 7238229.474 \approx 7.24x10^6##