On breaking a cubic solid (edge = 1 m) into fine cubic particles of...

For a cube of dimension $( 1m \times 1m \times 1m )$ , the surface area
$= ( 1 \times 1 ) \times 6$ faces
$= 6{m}^{2} \times {\left(\dfrac{1\mum}{1{0}^{- 6}m}\right)}^{2}$
$= 6 \times 1{0}^{12} ( \mum{)}^{2}$
The volume $\text{=}\text{1}{\text{m}}^{\text{3}} \text{\times}{\left(\dfrac{\text{1 \mu m}}{{\text{10}}^{- \text{6}} \text{m}}\right)}^{\text{3}}$
$= 1{0}^{18} ( \mum{)}^{3}$
For $1{0}^{18}$ cubes of dimension $( 1\mum \times 1\mum \times 1\mum )$ , the total surface area
$= 1{0}^{18} \times 1\mum \times 1\mum \times 6faces$
$= 6 \times 1{0}^{18} ( \mum{)}^{2}$
On breaking, the surface area becomes
$= \dfrac{6 \times 1{0}^{18} ( \mum{)}^{2}}{6 \times 1{0}^{12} ( \mum{)}^{2}}$ times
= 1000000 times
Hence n = 10