We have the expression for nuclear radius as:

$R =R_{0} A^{1 / 3}$

Where, $R _{0}=$ Constant.

$A =$ Mass number of the nucleus Nuclear matter density,

$\rho=\frac{\text {Mass of the micleus}}{\text {Volume of the nucleus}}$

Let $m$ be the average mass of the nucleus. Hence, mass of the nucleus $= mA$

$\rho=\frac{m A}{\frac{4}{3} \pi R^{3}}=\frac{3 m A}{4 \pi\left(R_{0} A^{1 / 3}\right)^{3}}=\frac{3 m A}{4 \pi R_{0}^{3} A}=\frac{3 m}{4 \pi R_{0}^{3}}$

Hence, the nuclear matter density is independent of $A$. It is nearly constant