Step 1: State the empirical relation.
The nuclear radius grows with mass number as \(R = R_0 A^{1/3}\), with \(R_0 \approx 1.2\ \text{fm}\). Cubing gives \(R^3 = R_0^3 A\), which is the key link we exploit.
Step 2: Density definition.
Density is mass per unit volume, \(\rho = M/V\). For a nucleus of mass number \(A\), the mass is \(M = Am\) (with \(m\) the average nucleon mass) and the volume is that of a sphere, \(V = \tfrac{4}{3}\pi R^3\).
Step 3: Insert the cubed radius.
Replace \(R^3\) by \(R_0^3 A\):
\[V = \frac{4}{3}\pi R_0^3 A\]
So the volume is directly proportional to \(A\).
Step 4: Form the ratio.
\[\rho = \frac{Am}{\tfrac{4}{3}\pi R_0^3 A} = \frac{3m}{4\pi R_0^3}\]
Because both mass and volume rise in exact proportion to \(A\), their ratio is constant. Numerically this comes out to about \(2.3\times10^{17}\ \text{kg/m}^3\) for every nucleus.
Step 5: Interpretation.
A heavier nucleus simply occupies a proportionally larger volume, so packing of nucleons stays the same. Nuclear matter has a universal density set only by \(m\) and \(R_0\).
\[\boxed{\rho = \frac{3m}{4\pi R_0^3}\ \text{is independent of } A}\]