The volume \( V \) of a hemisphere is given by the formula:

\(V = \frac{2}{3} \pi r^3. \)

To find the rate of change of volume \( \frac{dV}{dr} \) with respect to the radius \( r \), we need to take the derivative of the volume formula with respect to \( r \):

\(\frac{dV}{dr} = \frac{d}{dr} \left( \frac{2}{3} \pi r^3 \right). \)

Using the power rule for differentiation, where the derivative of \( x^n \) with respect to \( x \) is \( nx^{n-1} \), we get:

\(\frac{dV}{dr} = \frac{2}{3} \cdot 3 \pi r^2 = 2 \pi r^2. \)

Now, we are asked to find the rate of change of volume when \( r = 2 \):

\(\frac{dV}{dr} \bigg|_{r=2} = 2 \pi (2)^2 = 2 \pi \cdot 4 = 8 \pi. \)

So, the rate of change of the volume of the hemisphere with respect to its radius when \( r = 2 \) is \( 8 \pi \).

Therefore, the correct answer is:

A. \( 8 \pi \).