Explanation

(c)
Volume of one mole of silver atoms $= \frac{108}{10.5}{cm}^{3}/mol$
Volume of one silver atom $= \frac{108}{10.5} \times \frac{1}{6.022\ \times 10^{23}}\ {cm}^{3}$
So, $\frac{4}{3}\pi r^{3} = \frac{108}{10.5} \times \frac{1}{6.022\ \times 10^{23}} = 1.708 \times 10^{- 23}$
$\ \ r^{3} = 0.407 \times 10^{- 23}{cm}^{3} = 0.407 \times 10^{- 29}m^{3}$
Area of each silver atom,
$\ \ \pi r^{2} = \pi\left( 0.407 \times 10^{- 29}\ m^{3} \right)^{2/3}$
So, number of silver atoms in given area
$\ \ \ \ \ \ \ \ \ \ \ = \frac{10^{- 12}}{\left( 0.407 \times 10^{- 29}\ m^{3} \right)^{2/3}} = \frac{10^{8}}{\pi \times 2}$
$\ \ \ \ \ \ \ \ \ \ = 1.6 \times 10^{7} = y \times 10^{x}$
So, $\ \ x = 7$