Experiments show that the nuclear radius scales as $R = R_0\,A^{1/3}$ with $R_0 \approx 1.2\,\text{fm}$. A direct consequence of this is that

1. Nuclear volume: $V = \dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi R_0^3 A$ (using $R^3 = R_0^3 A$). 2. Nuclear mass: $M \approx A\,u$ (mass number times atomic mass unit). 3. Density: $\rho = M/V = \dfrac{A\,u}{\frac{4}{3}\pi R_0^3 A} = \dfrac{u}{\frac{4}{3}\pi R_0^3}$. 4. The $A$ in numerator and denominator CANCELS — the result depends only on universal constants $u$ and $R_0$. So nuclear density is the SAME for every nucleus (about $2.3\times 10^{17}\,\text{kg/m}^3$). 5. Sanity check (Example 13.1, NCERT): for iron $A=56$, the same formula gives $\rho \approx 2.3\times 10^{17}\,\text{kg/m}^3$. The same number emerges for any other $A$. _Source: NCERT Class 12 Physics Part 2, Ch 13, §13.3 (Size of the Nucleus) + Example 13.1, p. 4._