For a nucleus of mass number $A$ and radius $R$, the mass density of nucleus can be represented as

The mass density ( $\rho$) of a nucleus is defined as the ratio of its mass to its volume:
$\rho=$ Mass of nucleus / Volume of nucleus
For a spherical nucleus with radius R , the volume is:
$\text { Volume }=(4 / 3) \pi R^3$
The mass of the nucleus is related to the mass number A :
$\text { Mass }-\mathrm{A} \times \mathrm{m} \text { _u }$
(where m_u is the atomic mass unit, approximately $1.66 \times 10^{-27} \mathrm{~kg}$)
Therefore:
$\rho=\left(\mathrm{A} \times \mathrm{m} \_\mathrm{u}\right) /\left[(4 / 3) \pi \mathrm{R}^3\right]$
Now, there's an important relationship between the mass number A and nuclear radius R. Empirically, it has been found that:
$\mathrm{R}=\mathrm{Ro} * \mathrm{~A}^{\wedge}(1 / 3)$
Where Ro is a constant approximately equal to $1.2 \times 10^{-15} \mathrm{~m}(1.2$ fermi).
Substituting this into the density equation:
$\begin{aligned}<br/>& \rho=\left(\mathrm{A} \times \mathrm{m} \_\mathrm{u}\right) /\left[(4 / 3) \pi\left(\mathrm{Ro} \times \mathrm{A}^{\wedge}(1 / 3)\right)^3\right] \\ & \rho=(\mathrm{A} \times \mathrm{m} \text { _u }) /\left[(4 / 3) \pi \mathrm{Ro}^3 \times \mathrm{A}\right] \\ & \rho=\mathrm{m} \_\mathrm{u} /\left[(4 / 3) \pi \mathrm{Ro}^3\right]<br/>\end{aligned}$
This shows that the nuclear density is approximately constant for all nuclei, regardless of their mass number. This is a fundamental property of nuclear matter known as nuclear saturation density.
The numerical value is:
$\rho=2.3 \times 10^{17} \mathrm{~kg} / \mathrm{m}^3$