Hello Elizabeth,
I prefer to work problems in general terms, and derive a formula we may then plug our data into. Let's let the radius of the cylindrical tank be $r$, the height of the tank be $h$ and the additional distance above the tank the fluid must be pumped be $q$. All measures are in meters.
Let's then imagine the surface of the fluid is initially a distance $h_1$ from the top of the tank, and we wish to pump out fluid until the surface is a distance $h_2$ from the top of the tank. We will let $W$ be the amount of work required to accomplish this task.
We will use a vertical $y$-axis along which to integrate. Now, if we imagine slicing the cylinder of water we wish to remove into disks, we may state, using work is force times distance for a constant force:
$$dW=Fd=mgd$$
The force exerted is equal to the weight of the slice, which is the product of the mass and the acceleration due to gravity.
The mass $m$ of the slice is the product of the mass density $\rho$ (in $$\frac{\text{kg}}{\text{m}^3}$$) and the volume $$V=\pi r^2\,dy$$ of the slice, i.e.:
$m=\rho\pi r^2\,dy$
The distance the slice must be vertically moved against gravity is:
$d=q+y$
Putting it all together, we have:
$$dW=\rho\pi g r^2(q+y)\,dy$$
Now, to find the total work done, we may sum the work differentials through integration:
$$W=\rho\pi g r^2\int_{h_1}^{h_2}(q+y)\,dy$$
Applying the anti-derivative form of the FTOC, we find:
$$W=\rho\pi g r^2\left[qy+\frac{1}{2}y^2 \right]_{h_1}^{h_2}$$
$$W=\rho\pi g r^2\left(\left(qh_2+\frac{1}{2}h_2^2 \right)-\left(qh_1+\frac{1}{2}h_1^2 \right) \right)$$
$$W=\rho\pi g r^2\left(q(h_2-h_1)+\frac{1}{2}(h_2^2-h_1^2) \right)$$
$$W=\rho\pi g r^2(h_2-h_1)\left(q+\frac{1}{2}(h_2+h_1) \right)$$
$$W=\frac{1}{2}\rho\pi g r^2(h_2-h_1)\left(h_1+h_2+2q \right)$$
Now we may plug in the given/known data
Water is generally taken to have a mass density of:
$$\rho=1000\,\frac{\text{kg}}{\text{m}^3}$$
I will assume we are to use:
$$g=9.8\,\frac{\text{m}}{\text{s^2}}$$
We are given:
$r=3\text{ m},\,q=2\text{ m},\,h=8\text{ m}$
For part (A), we have:
$$h_1=0\text{ m},\,h_2=\frac{h}{2}=4\text{ m}$$
Hence:
$$W_A=1411200\pi\text{ J}$$
For part (B), we have:
$$h_1=\frac{h}{2}\,\text{m}=4\text{ m},\,h_2=h=8\text{ m}$$
Hence:
$$W_B=2822400\pi\text{ J}=2W_A$$
(C) If we interpret the amount of work done as the area of a trapezoid, since the integrand is a linear function, we find that:
$$W_A=k\cdot\frac{1}{2}(6+2)4=16k$$
$$W_B=k\cdot\frac{1}{2}(10+6)4=32k=2W_A$$
