On breaking a cubic solid (edge = 1 m) into fine cubic particles of edge 1 $\mathrm{\mu }\mathrm{m}$ , surface becomes $\mathrm{n}\times 100000$ . The value of n is?

For a cube of dimension $(1\mathrm{m}\times 1\mathrm{m}\times 1\mathrm{m})$ , the surface area
$=(1\times 1)\times 6$ faces
$=6{\mathrm{m}}^2\times {\left(\frac{1\mathrm{\mu }\mathrm{m}}{10^{-6}\mathrm{m}}\right)}^2$
$=6\times 10^{12}(\mathrm{\mu }\mathrm{m}{)}^2$
The volume $\text{=}\text{1}{\text{m}}^{\text{3}}\text{×}{\left(\frac{\text{1μm}}{{\text{10}}^{-\text{6}}\text{m}}\right)}^{\text{3}}$
$=10^{18}(\mathrm{\mu }\mathrm{m}{)}^3$
For $10^{18}$ cubes of dimension $(1\mathrm{\mu }\mathrm{m}\times 1\mathrm{\mu }\mathrm{m}\times 1\mathrm{\mu }\mathrm{m})$ , the total surface area
$=10^{18}\times 1\mathrm{\mu }\mathrm{m}\times 1\mathrm{\mu }\mathrm{m}\times \mathrm{6 }\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{s}$
$=6\times 10^{18}(\mathrm{\mu }\mathrm{m}{)}^2$
On breaking, the surface area becomes
$=\frac{6\times 10^{18}(\mathrm{\mu }\mathrm{m}{)}^2}{6\times 10^{12}(\mathrm{\mu }\mathrm{m}{)}^2}$ times
= 1000000 times
Hence n = 10