At what temperature will oxygen molecules have the same root mean square speed as hydrogen molecule at 200K?

We know that, ${\mathrm{V}}_{\text{ms }}=\sqrt{(3\mathrm{kT}/\mathrm{m})}$
     $\therefore {\left({\mathrm{v}}_{\mathrm{rms}}\right)}_{\mathrm{o}}=\sqrt{\left(3{\mathrm{kT}}_{\mathrm{o}}/{\mathrm{m}}_{\mathrm{o}}\right)}$
and   ${\left({\mathrm{V}}_{\mathrm{rms}}\right)}_{\mathrm{h}}=\sqrt{\left(3{\mathrm{kT}}_{\mathrm{h}}/{\mathrm{m}}_{\mathrm{h}}\right)}$
The value of ${\mathrm{v}}_{\mathrm{rms}}$ will be the same, if $\frac{{\mathrm{T}}_0}{{\mathrm{m}}_0}=\frac{{\mathrm{T}}_{\mathrm{h}}}{{\mathrm{m}}_{\mathrm{h}}}$
or    ${\mathrm{T}}_{\mathrm{o}}=\frac{{\mathrm{m}}_0}{{\mathrm{m}}_{\mathrm{h}}}\times {\mathrm{T}}_{\mathrm{h}}=16\times 200=3200\mathrm{K}={2927}^{\circ }\mathrm{C}$