Answer
$583\;\rm m/s$
Work Step by Step
We know that the kinetic energy of a monatomic gas particle is entirely translational kinetic energy and hence the thermal energy of a monatomic gas of $N$ atoms is given by
$$E_{th}=NK_{avg}$$
where $K_{avg}$ is the average kinetic energy of one atom (or molecule).
$$E_{th}=N(\frac{1}{2}mv_{avg}^2)$$
Hence,
$$v_{avg}=\sqrt{\dfrac{2E_{th}}{mN}}\tag 1$$
Now we need to find the number of atoms in the 10-g sample of neon where $N=nN_{\rm A}$ and $n=m'/M$ where $m'$ is the mass of the sample and $M$ is the atomic mass of the neon.
Hence,
$$N=\dfrac{m'}{M_{Ne}}N_A$$
Plugging into (1);
$$v_{avg}=\sqrt{\dfrac{2E_{th}M_{Ne}}{mm'N_{\rm A}}} $$
Plugging the known; and recall that $m$ is the mass of the neon atom which is given by $m=(20.2\times 1.661\times 10^{-27})$
$$v_{avg}=\sqrt{\dfrac{2(1700)(20.2)}{(20.2\times 1.661\times 10^{-27})(10) (6.022\times 10^{23})}} $$
$$v_{avg}=\color{red}{\bf 583}\;\rm m/s$$
