The mean kinetic energy of the molecules of an ideal gas at $399^{\circ} \mathrm{C}$ is ' E '. The temperature at which the mean kinetic energy of its molecules will be ' $E / 2$ ', is

The mean kinetic energy $E$ of an ideal gas is proportional to its absolute temperature, given by $E = \frac{3}{2} k_B T$, where $k_B$ is Boltzmann's constant.

Given the initial temperature $399^\circ \mathrm{C}$, convert to Kelvin: $T_1 = 399 + 273 = 672 \mathrm{K}$.

At temperature $T_1$, the energy is $E$; we seek the temperature $T_2$ for energy $E/2$.
Using the proportionality $E \propto T$, we have $\frac{E_2}{E_1} = \frac{T_2}{T_1}$.
Substituting $E_2 = E/2$ and $E_1 = E$ yields $\frac{E/2}{E} = \frac{T_2}{672 \mathrm{K}}$, so $\frac{1}{2} = \frac{T_2}{672 \mathrm{K}}$.

Solving, $T_2 = \frac{672 \mathrm{K}}{2} = 336 \mathrm{K}$.
Converting to Celsius: $T_2 = 336 - 273 = 63^\circ \mathrm{C}$.

Final answer: $\boxed{\text{D}}$