The radius $ R $ of a nucleus is given by the empirical relation:

$$ R = r_0 A^{\frac{1}{3}}, $$

where $ r_0 $ is a constant and $ A $ is the mass number.

The volume $ V $ of the nucleus is approximately that of a sphere:

$$ V = \frac{4}{3}\pi R^3. $$

Substituting the expression for $ R $ into the volume formula:

$$ V = \frac{4}{3}\pi \left(r_0 A^{\frac{1}{3}}\right)^3 = \frac{4}{3}\pi r_0^3 A. $$

The mass density $ \rho $ is defined as the mass (which is proportional to $ A $) divided by the volume:

$$ \rho = \frac{A}{V} = \frac{A}{\frac{4}{3}\pi r_0^3 A} = \frac{1}{\frac{4}{3}\pi r_0^3}. $$

Notice that $ A $ cancels out in the numerator and denominator, so the density is:

$$ \rho = \text{constant}, $$

meaning it is independent of the mass number $ A $.

Thus, the correct option is:

Option B: Independent of A.