The tank is a truncated circular cone with a base of radius $r=3\,$m, a depth of $4\,$m and top radius of $r=4\,$m. Let $y$ denote the vertical distance from the bottom of the tank. Then $r=3+\frac{1}{4}y$ is the radius of the slice of water lying $y$ meters above the bottom of the tank.
Think of that slice of water as being solid rather than liquid, as if it were frozen. The amount of work done in pumping the water to a height of $5\,$m above the base is the same as if one had to move all those frozen slices to that height.
The volume of each slice is $\pi r^2\,dy$ with $dy$ representing the thickness. The mass of the slice (in the mks system) is found by multiplying the density $\rho=1000\,$ kg/m$^3$ times the volume, so
\begin{equation}
M=1000\pi r^2\,dy=1000\pi\left(3+\frac{1}{4}y\right)^2dy
\end{equation}
Each of these slices of water must be moved upward a distance of $D=5-y$ meters against a gravitational force of $g=9.8\,$m/sec$^2$ resulting in the work for the slice at $y$ being
\begin{equation}
W_y=9800\pi\left(3+\frac{1}{4}y\right)^2(5-y)\,dy
\end{equation}
The total work to remove all the slices of water between $y=0$ and $y=4$ is
\begin{equation}
W=\int_{0}^{4}9800\pi\left(3+\frac{1}{4}y\right)^2(5-y)\,dy
\end{equation}
The rest is routine since the integrand is simply a third degree polynomial. The units will be joules.
