$\mathrm{d}=\frac{\mathrm{m}}{\mathrm{v}}\Rightarrow 10.5\mathrm{g}/\mathrm{cc}$

Number of atoms of Ag in 1 cc $\Rightarrow \sqrt[3]{\frac{10.5}{108}\times {\mathrm{N}}_{\mathrm{A}}}$

In 1cm, number of atoms of  $\mathrm{Ag}={(\frac{10.5}{108}\times {\mathrm{N}}_{\mathrm{A}})}^{2/3}$

In 10^–12 m² or 10^–8 cm², number of atoms of $\mathrm{Ag}={\left(\frac{10.5}{108}{\mathrm{N}}_{\mathrm{A}}\right)}^{2/3}\times {10}^{-8}=\left(\frac{10.5\times 6.02\times {10}^{23}}{108}\right)\times {10}^{-8}$

= 1.5 × 10⁷

Thus, x = 7