Step-by-step Solution:

1. Identify the formula for the volume of a cone. The volume ($V$) of a cone is given by the formula: $V = \frac{1}{3} \pi r^2 h$ where $r$ is the radius of the base and $h$ is the height. Since the diameter $D$ is given, we can express the radius as $r = D/2$. Substituting this into the volume formula: $V = \frac{1}{3} \pi \left(\frac{D}{2}\right)^2 h = \frac{1}{3} \pi \frac{D^2}{4} h = \frac{\pi}{12} D^2 h$

2. Determine the formula for maximum percentage error. For a quantity $Q$ that depends on variables $A$, $B$, etc., as $Q = k A^a B^b ...$, the maximum fractional error in $Q$ is given by: $\frac{\Delta Q}{Q} = a \frac{\Delta A}{A} + b \frac{\Delta B}{B} + ...$ In our case, $V = (\pi/12) D^2 h^1$. The constant $\pi/12$ has no error. The variables are $D$ and $h$. The powers are $a=2$ for $D$ and $b=1$ for $h$. So, the maximum fractional error in the volume is: $\frac{\Delta V}{V} = 2 \frac{\Delta D}{D} + 1 \frac{\Delta h}{h}$ The maximum percentage error is obtained by multiplying the fractional error by 100: $\% \text{error in } V = \left( 2 \frac{\Delta D}{D} + \frac{\Delta h}{h} \right) \times 100\%$

3. Identify the given values and the error in measurements.

   • Measured diameter, $D = 20.0 \mathrm{~cm}$.
   • Measured height, $h = 20.0 \mathrm{~cm}$.
   • Least count of the scale = $2 \mathrm{~mm}$. The maximum possible error in a measurement is equal to the least count of the measuring instrument.
   • Error in diameter, $\Delta D = 2 \mathrm{~mm}$.
   • Error in height, $\Delta h = 2 \mathrm{~mm}$.

4. Ensure consistent units. The measurements are in cm, while the error is in mm. We need to convert them to the same unit. Let's convert mm to cm ($1 \mathrm{~cm} = 10 \mathrm{~mm}$).

   • $\Delta D = 2 \mathrm{~mm} = 0.2 \mathrm{~cm}$.
   • $\Delta h = 2 \mathrm{~mm} = 0.2 \mathrm{~cm}$.

5. Calculate the fractional errors in diameter and height.

   • Fractional error in diameter: $\frac{\Delta D}{D} = \frac{0.2 \mathrm{~cm}}{20.0 \mathrm{~cm}} = \frac{2}{200} = \frac{1}{100} = 0.01$
   • Fractional error in height: $\frac{\Delta h}{h} = \frac{0.2 \mathrm{~cm}}{20.0 \mathrm{~cm}} = \frac{2}{200} = \frac{1}{100} = 0.01$

6. Calculate the maximum percentage error in the volume. Substitute the fractional errors into the formula from Step 2: $\% \text{error in } V = \left( 2 \times (0.01) + 1 \times (0.01) \right) \times 100\%$ $\% \text{error in } V = (0.02 + 0.01) \times 100\%$ $\% \text{error in } V = 0.03 \times 100\%$ $\% \text{error in } V = 3\%$

The maximum percentage error in the determination of the volume is 3.