\[ \boxed{Q \approx 19.8 \, \text{cm}^3/\text{s}} \]
An alternative route is to derive the exit speed from the work-energy theorem applied to the falling water column itself, treating the water leaving the hole as if it were a particle that has effectively "fallen" through the height \( h \), then check the answer using unit consistency.
A particle falling freely from rest through a height \( h = 20 \, \text{m} \) under gravity \( g = 9.81 \, \text{m/s}^2 \) gains kinetic energy equal to the work done by gravity:
\[ \frac{1}{2}v^2 = gh \implies v = \sqrt{2gh} = \sqrt{2(9.81)(20)} \approx 19.81 \, \text{m/s} \]
This is the same speed a free-falling drop would reach after dropping \( 20 \, \text{m} \), which is physically sensible since the pressurized water column above the hole is effectively converting that same gravitational potential energy into the kinetic energy of the jet.
The flow rate is then area times speed, with the area carefully kept in consistent units:
\[ A = 1 \, \text{mm}^2 = 0.01 \, \text{cm}^2, \quad v = 19.81 \, \text{m/s} = 1981 \, \text{cm/s} \]
\[ Q = Av = 0.01 \times 1981 \approx 19.8 \, \text{cm}^3/\text{s} \]
Checking against the options:
The free-fall analogy confirms option D.
Therefore, the correct answer is 19.8 cm³/s.