Answer
$v_p=5.0 \times 10^5 \mathrm{~m} / \mathrm{s}$.
Work Step by Step
$
v_p=\sqrt{\frac{2 R T}{M}}=\sqrt{\frac{2 R T}{m N_A}}=\sqrt{\frac{2 k T}{m}} .
$
With $T=1.5 \times 10^7 \mathrm{~K}$ and $m=1.67 \times 10^{-27} \mathrm{~kg}$, this yields $v_p=5.0 \times 10^5 \mathrm{~m} / \mathrm{s}$.
