Why is the van der Waals coefficient b equal to four times the volume of the particle?
While most everything the previous answer states is correct, I would point out that taking four times the volume of a single particle has nothing to do with experiment and arises mathematically.
In deriving the VDW equation, the particles are still assumed to be hard spheres, but this assumption is corrected for with the parameter $a$.
The hard sphere approximation forbids that two particles penetrate each other's radii. Thus, we find that two spheres in closest contact are surrounded by a sphere of radius $2r$ (or the diameter of one of the original spheres).

(source: nyu.edu)
Thus, the volume excluded by the particles from the larger sphere surrounding the two spheres shown is
$$b' = \frac{4}{3} \pi d^3 = 8 \cdot \frac{4}{3} \pi r^3$$
Thus, the excluded volume per particle $b$ is $b'/2$ or,
$$b=4 \cdot \frac{4}{3}\pi r^3$$
which, as you point out, is four times the volume of a single particle.
The interesting thing about this is that it does not represent the actual value of $b$ for any given atom, but represents the upper bound of $b$ for any given atom. What I mean by that is, $b$ could very well be correct by calculating four times the volume, but often experiment will show that it less than the calculated value of $b$ because atoms are not hard spheres.
For instance, using helium, which is the closest we'll get to a hard sphere:
$$b_{\ce{He},\mathrm{calc}} = 4 \cdot \frac{4}{3} \pi (140\ \mathrm{pm})^3 \cdot N_\mathrm{A} = 0.02767\ \mathrm{L\ mol^{-1}}$$
while,
$$b_{\ce{He},\mathrm{exp}} = 0.0238\ \mathrm{L\ mol^{-1}}$$
So, the experimental value of $b$ is indeed smaller, but the calculated value gives a rough idea.