Radius of the atomic nucleus in terms of mass number is given by,

$R=R_0{A}^{\frac{1}{3}}.$ Clearly as mass number increases, volume and mass of the nucleus changes.

Similarly, as the size of the nucleus changes, binding energy also changes.

For density we can write,

$$\begin{gathered} \frac{\text{mass of nucleus}}{\text{volume of nucleus}} \\ =\frac{mA}{{\displaystyle \frac{4}{3}\pi R^3}} \\ =\frac{{\displaystyle mA}}{{\displaystyle \frac{4}{3}\pi {\left(R_0\right)}^{3}A}} \\ =\frac{3m}{4\pi {\left(R_0\right)}^3} \\ =\text{constant} \end{gathered}$$