Let's use letter c for cup and letter e for hemisphere. The information above can be rewritten as \(r_c = \frac{1}{2}r_e\) and \(h_c = 2r_c\).

The volume of the cup, given that it is a cylinder is equal to \(\pi r_c^2 h_c\), while the volume of the hemisphere is half the volume of a sphere, i.e. \(\frac{1}{2}(\frac{4}{3}\pi r_e^3) = \frac{2}{3}\pi r_e^3\). Substituting in the formula for the volume of the cup, using the information given, leads to \(\pi \frac{1}{4}r_e^2 r_e = \frac{1}{4} \pi r_e^3\).

Now we can multiply the volume of the cup by our choices to get the one which does not exceed the volume of the hemisphere.

If we use 2 cups, the volume becomes \(\frac{1}{2} \pi r_e^3\), which is less than \(= \frac{2}{3}\pi r_e^3\).
If we use 3 cups, the volume becomes \(\frac{3}{4} \pi r_e^3\), which is higher than \(= \frac{2}{3}\pi r_e^3\).

Answer A