To evaluate the assertion and reason provided, we need to consider the formula for the density of a nucleus. The density $ \rho $ can be expressed as follows:

$ \rho = \frac{M}{V} = \frac{m_n \times A}{\frac{4}{3} \pi R^3} $

Here, $ m_n $ represents the nucleon mass, $ A $ is the mass number, and $ R $ is the radius of the nucleus. According to the formula for the radius:

$ R = R_0 A^{1/3} $

This indicates that the radius $ R $ is proportional to $ A^{1/3} $. By substituting the expression for $ R $ in terms of $ A $ back into the formula for density, we get:

$ \rho = \frac{m_n \times A}{\frac{4}{3} \pi (A^{1/3} R_0)^3} $

Simplifying the equation shows that the $ A $ terms cancel out:

$ \rho = \frac{m_n}{\frac{4}{3} \pi R_0^3} $

This reveals that the nuclear density $ \rho $ is approximately constant and independent of the mass number $ A $. Therefore, the densities of copper $(^{64}_{29} \text{Cu})$ and carbon $(^{12}_{6} \text{C})$ nuclei are effectively the same because the density formula results in a similar value regardless of the specific mass number.