d
Kinetic energy of each molecule,
$\mathrm{K} \mathrm{E}=\frac{3}{2} \mathrm{K}_{\mathrm{B}} \mathrm{T}$
In the given problem,
Temperature, $\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{K}$
Height attained by the gas molecule, $h=?$
$\mathrm{KE}=\frac{3}{2} \mathrm{K}_{\mathrm{B}}(273)=\frac{819 \mathrm{K}_{\mathrm{B}}}{2}$
$\mathrm{K.E}=\mathrm{PE}$
$ \Rightarrow \frac{{819{K_B}}}{2} = Mgh$
$h=\frac{{819{k_B}}}{{2Mg}}$