Explanation

(d)
Number of unit cells $= \frac{mass\ of\ metal\ }{mass\ of\ one\ unit\ cell\ }$
Given, edge length of unit cell $= 2\mathring{\mathrm{A}} = 2 \times 10^{- 8}\ cm$
Mass of metal $= 200\ g$
Density of metal $= 2.5\ g\ {cm}^{- 3}$
Volume of unit cell $= (edge\ length)^{3} = \left( 2 \times 10^{- 8} \right)^{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 8 \times 10^{- 24}\ {cm}^{3}$
Mass of one unit cell $= volume\ \times density$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 8 \times 10^{- 24} \times 2.5$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 20 \times 10^{- 24}$
$\therefore$ No. of unit cells in 200 g metal $= \frac{mass\ of\ metal\ }{mass\ of\ one\ unit\ cell\ }$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{200}{20 \times 10^{- 24}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10 \times 10^{24} = 1.0 \times 10^{25}$