I found it easier to translate the sphere from sitting on top of the table to being sliced in half by the table. Then the horizontal plane that cuts the sphere is 8 cm above the table. I agree with your calculation of 2680.8 cm^3. To start, I drew a circle centered at (0,0) with radius 16:
[math]x^2+y^2=16^2\\ x=\sqrt{16^2-y^2}[/math]
The plane slices the sphere, and the intersection gives a circle. At a height [imath]y=8[/imath], the radius is given by [imath]x=\sqrt{16^2-8^2}[/imath]. Area of that circle is [imath]\pi r^2=\pi ( 16^2-8^2 )[/imath]. Thicken the circle into a 3D disc of volume [imath]\pi (16^2-8^2)dy[/imath]. Add up all of the disc volumes at each [imath]y[/imath] from 8 cm to the top of the sphere at 16 cm using integration:
[math]\int_{y=8}^{16}\pi(16^2-y^2)dy\approx 2680.8 \text{ cm}^3[/math]
To find the work of pumping the liquid in the sphere below the plane to just above the plane, I set up an integral as well. Each horizontal thin disc of liquid has a volume, calculated similar to the above, that is converted to mass by the density [imath]\rho[/imath]. Multiply mass by [imath]g[/imath] to get force. Then multiply force by distance to get work. Each thin disc of liquid is pumped a vertical distance of [imath]8-y[/imath], which works for negative values of [imath]y[/imath] too. I'll let you set up the definite integral, but it should look something like
[math]\int_{-16}^8\rho\cdot(\text{volume of disc})\cdot g\cdot (8-y)[/math]
Only tricky part is to include the conversion factors within the integral so that you get units of Joules. I get about [imath]1.882 g\text{ J}[/imath].
