#### Answer

$v = 1360~m/s$

#### Work Step by Step

We can find the kinetic energy of one helium atom.
$KE = \frac{3700~J}{6.02\times 10^{23}~atoms}$
$KE = 6.146\times 10^{-21}~J$
We can find the speed of each helium atom.
$\frac{1}{2}mv^2 = KE$
$v^2 = \frac{2~KE}{m}$
$v = \sqrt{\frac{2~KE}{m}}$
$v = \sqrt{\frac{(2)(6.146\times 10^{-21}~J)}{6.68\times 10^{-27}~kg}}$
$v = 1360~m/s$