Finding the Radius of an Electron, using the known magnetic moment, and assuming that the little guy is spinning at the speed of light.

$$\require{physics}$$ $$ \mu = \frac{ -e }{ 2 m } L $$

Where:
mu = the magnetic moment of an electron
e = the charge of an electron
m = the mass of an electron
L = the angular momentum of an electron
r = the radius of an electron
v = the velocity at which an electron is rotating

$$ L = I \omega $$ $$ L = r^2 m \left( \frac{v}{r} \right) $$ $$ L = r m v $$

Substitute for L

$$ \mu = \frac{-e}{2m} L $$ $$ \mu = \frac{-e}{2m} r m v $$ $$ \mu = \frac{-e}{2} r v $$ $$ \mu = \frac{-e r v }{2} $$ $$ r = \frac{2 \mu} {-e v} $$

We know the values of mu and e. Let us assume, as Robin suggests, that v = c = 3*10^8 m/s and solve for r, the radius of the electron.

$$ v = 3 * 10^8 $$ $$ r = \frac{2 \mu} {-e v} $$ $$ r = \frac{2 (-9.284765 *10^{-24}) }{ (-1.602177 * 10^{-19} )( 3*10^{8} )} $$ $$ r = 3.863395* 10^{-13} $$

This is even bigger than the "Classical Electron Radius", whose CODATA value is:

$$ r_e = 2.817 940 3262 * 10^{-15} $$

The "Classical Electron Radius" is 2.5 times the size of a Proton! Did I make an error in my calculations? If we want to decrease our value for r, then we must increase v to some value greater than the speed of light.