As we know, $\text{density of nuclei}=\frac{\text{ mass of nuclei }}{\text{ volume of nuclei }}$

$\Rightarrow \rho =\frac{1.6\times {10}^{-27}A}{\frac{4}{3}\pi {\left(1.5\times {10}^{-15}\right)}^{3}A}$

$$\begin{gathered} =\frac{1.6\times {10}^{-27}}{14.14\times {10}^{-45}} \\ =0.113\times {10}^{18} \mathrm{kg} {\mathrm{m}}^{-3} \end{gathered}$$

We also know that, density of water is ${\rho }_{\mathrm{w}}={10}^3 \mathrm{kg} {\mathrm{m}}^{-3}$.

Hence,
$\frac{\rho }{{\rho }_w}=11.31\times {10}^{13}$.

Therefore, $n\approx 11$