Here, average radius of sodium atom,$r = 2.5 \text{\AA} = 2.5 \times 10^{-10} \text{m}$ ∴ Volume of sodium atom $= \frac{4}{3} \pi r^{3}$ $= \frac{4}{3} \times 3.14 \times (2.5 \times 10^{-10})^{3}$ $= 65.42 \times 10^{-30} \text{m}^{3}$ Mass of a mole of sodium $=23$ gram $= 23 \times 10^{-3} \text{kg}$ Also we know that each mole contains $6.023 \times 10^{23}$ atoms, hence the massof sodium atom, $M = \frac{23 \times 10^{-3}}{6.023 \times 10^{23}} \text{kg}$ $= 3.82 \times 10^{-26} \text{kg}$ $\therefore$ Average mass density of sodium atom. $\rho = \frac{M}{V}$ $= \frac{3.82 \times 10^{-26}}{65.42 \times 10^{-30}} \text{kg} \text{m}^{-3}$ $= 0.58 \times 10^{3} \text{kg} \text{m}^{-3}$ Density of sodium in crystalline phase $= 970 k \text{gm}^{-3} = 0.970 \times 10^{3} \text{kg} \text{m}^{-3}$ $\therefore \frac{\text{Average mass density of sodiumatom}}{\text{Density of sodium in crystalline phase}}$$= \frac{0.58 \times 10^{3}}{0.970 \times 10^{3}}$$=0.66$Yes, both densities are of the same order of magnitude, i.e. of the order of10$^{3}$.This is because in the solid phase atoms are tightly packed.