Answer
$3.5\times10^{-9}m/s$
Work Step by Step
The average translational kinetic energy of a gas molecule is $\frac{3}{2}kT$. Set this equal to the KE of a paper clip.
$$\frac{1}{2}mv^2=\frac{3}{2}kT$$
$$v=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3(1.38\times10^{-23}J/K)(273+22K)}{ 1.0\times10^{-3}kg}}\approx3.5\times10^{-9}m/s$$
