The order of magnitude of density of uranium nucleus is: ($m_p$= 1.67 x 10^{- 21} kg)  

If $A$ is the atomic number, the mass of uranium nucleus is:

 $m=\left(1.67\times {10}^{-27}\right) \mathrm{A}$kg and its volume is: 

$V=\frac{4}{3}\pi r^3=\frac{4\pi }{3}\times {\left\{1.25\times {10}^{-15}{A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right\}}^3$

    $\simeq (8.2\times {10}^{-45} A) {\mathrm{m}}^3$             $\left[\because \mathrm{r}={\mathrm{r}}_0{\mathrm{A}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}; {\mathrm{r}}_{0=}1.25\times {10}^{-15} \mathrm{m}\right]$

$\therefore$Density = $\frac{\mathrm{m}}{\mathrm{V}}=\frac{1.67\times {10}^{-27}\mathrm{A} \mathrm{kg}}{8.2\times {10}^{-45}\mathrm{A} {\mathrm{m}}^3}$

                            $\simeq 2\times {10}^{17 }$$\raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$m^3$}\right.$