The radius of both Deuterium and Tritium are approximately 2.0 fm. The temperature to which the gas must be heated to initiate the fusion reaction is

$\begin{array}{l}{\mathrm{E}}_{\mathrm{avg}}=\frac{3}{2}{\mathrm{K}}_{\mathrm{B}}\mathrm{T};{\mathrm{E}}_{\mathrm{avg}}= \mathrm{E}.\mathrm{P}.\mathrm{E}\\ \frac{3}{2}{\mathrm{K}}_{\mathrm{B}}\mathrm{T}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\cdot \frac{{\mathrm{q}}_1{\mathrm{q}}_2}{2\mathrm{r}}\\ \mathrm{T}=\frac{2}{3{\mathrm{K}}_{\mathrm{B}}}\times \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{2\mathrm{r}}\end{array}$
K_B = 1.38×10^–23J/K(Boltzman's const)
$\begin{array}{l}{\mathrm{q}}_1={\mathrm{q}}_2=1.6\times {10}^{-19}\mathrm{C};\mathrm{r}=2\mathrm{fm}=2\times {10}^{-15}\mathrm{m}\\ \therefore \mathrm{T}=\frac{2}{3}\times \frac{9\times {10}^9\times {\left(1.6\times {10}^{-19}\right)}^2}{1.38\times {10}^{-23}\times 2\times 2\times {10}^{-15}}\\ \therefore \mathrm{T}=2.78\times {10}^9\mathrm{K}\end{array}$