Explanation

(d)
Volume of the gold dispersed in one litre water
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{mass}{density} = \ \frac{1.9 \times 10^{- 4}g}{19g\ {cm}^{- 3}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 \times 10^{- 5}{cm}^{- 3}$
Radius of gold sol particle =10 nm
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10 \times 10^{- 9}m = 10 \times 10^{- 7}cm$
Volume of the gold sol particle
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{4}{3}\pi r^{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{4}{3} \times \frac{22}{7} \times \left( 10^{- 6} \right)^{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 4.19 \times 10^{- 18}{cm}^{3}$
Number of gold sol particle in 1$\times 10^{- 5}{cm}^{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1 \times 10^{- 5}}{4.19 \times 10^{- 18}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.38 \times 10^{12}$
Number of gold sol particle in one mm^{3}
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{2.38 \times 10^{12}}{10^{6}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.38 \times 10^{6}$