Finding root-mean-square speed of neon using kinetic energy Find the root-mean-square speed of $\ce{Ne}$ atoms at the temperature at which their kinetic energy is $\pu{6.24 kJ mol-1}.$

I tried using the kinetic energy formula
$$\mathrm{KE} = \frac{mv^2}{2},$$
but I don't really understand how to achieve the necessary values.
I tried to do it by converting mass of one atom to mass of a mole since the given energy is per mole, but I'm still not getting the answer. I understand that I need to do $\displaystyle\sqrt{\frac{3RT}{M}},$ where $R= \pu{8.3145 J mol^-1 K^-1}$ and $M = 20,$ but how would I get the temperature?
As hinted in the comments by @AngusTheMan we use the equation
$$\mathrm{KE_{avg}} = \frac{3}{2}kT,$$
but ultimately we are looking for the $v_\mathrm{rms}$, so we'll also use
$$v_\mathrm{rms} = \sqrt{\frac{3RT}{M}}.$$
You've been given your $\mathrm{KE_{avg}},$ $R$ and $k$ are constants, and $M$ is the mass of 1 mole of $\ce{Ne}$. We can set all of these equal and the solution will come from
$$v_\mathrm{rms} = \sqrt{\frac{3 \cdot R\cdot\mathrm{KE_{avg}}\cdot\frac{2}{3k}}{M}}.$$