Explanation

(d)
Edge length of the unit cell $= 2\ \mathring{\mathrm{A}} = 2 \times 10^{- 8}cm$
Volume of the unit cell $= \left( 2 \times \ 10^{- 8} \right)^{3}\ cm^{3}$
$= 8\ \times \ 10^{- 24}\ cm^{3}$
Mass of unit cell = volume $\times$ density
$= 8\ \times \ 10^{- 24}\ \times \ 2.5\ g$
No. of unit cells in 200 g of the metal
$= \frac{mass\ of\ metal}{mass\ of\ unit\ cell} = \frac{200}{8\ \times \ 10^{- 24} \times 2.5}$
$= 1\ \times \ 10^{25}$