If $r$ is the radius of the atom then the volume of each atom will be $(4 / 3) \pi r^{3}$
Volume of all the substance will be $=(4 / 3) \pi r^{3} \times N=M / \rho$
$M=$ atomic mass of the substance
$\rho=$ density of the substance
We know, One mole of the substance has $6.023 \times 10^{23}$ atoms
$r=\left(3 M / 4 \pi \rho \times 6.023 \times 10^{23}\right)^{1 / 3}$
For carbon, $M=12.01 \times 10^{-3} \mathrm{~kg}$ and $\rho=2.22 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$
$R=\left(3 \times 12.01 \times 10^{-3} / 4 \times 3.14 \times 2.22 \times 10^{3} \times 6.023 \times 10^{23}\right)^{1 / 3}$
$=\left(36.03 \times 10^{-3} / 167.94 \times 10^{26}\right)^{1 / 3}$
$1.29 \times 10^{-10} \mathrm{~m}=1.29 \AA$
For gold, $M=197 \times 10^{-3} \mathrm{~kg}$ and $\rho=19.32 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$
$R=\left(3 \times 197 \times 10^{-3} / 4 \times 3.14 \times 19.32 \times 10^{3} \times 6.023 \times 10^{23}\right)^{1 / 3}$
$=1.59 \times 10^{-10} \mathrm{~m}=1.59 \AA$
For lithium, $M=6.94 \times 10^{-3} \mathrm{~kg}$ and $\rho=0.53 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
$R=\left(3 \times 6.94 \times 10^{-3} / 4 \times 3.14 \times 0.53 \times 10^{3} \times 6.023 \times 10^{23}\right)^{1 / 3}$
$=1.73 \times 10^{-10} \mathrm{~m}=1.73 \AA$
For nitrogen (liquid), $M=14.01 \times 10^{-3} \mathrm{~kg}$ and $\rho=1.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
$R=\left(3 \times 14.01 \times 10^{-3} / 4 \times 3.14 \times 1.00 \times 10^{3} \times 6.023 \times 10^{23}\right)^{1 / 3}$
$=1.77 \times 10^{-10} \mathrm{~m}=1.77 \AA$
For fluorine (liquid), $M=19.00 \times 10^{-3} \mathrm{~kg}$ and $\rho=1.14 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
$R=\left(3 \times 19 \times 10^{-3} / 4 \times 3.14 \times 1.14 \times 10^{3} \times 6.023 \times 10^{23}\right)^{1 / 3}$
$=1.88 \times 10^{-10} \mathrm{~m}=1.88 \AA$