You have found an elaborate way of calculating $2\pi \alpha/ \ln 2 \approx 0.0661658$. Here, $\alpha \approx 1/137$ represents the fine-structure constant.
The points to note is that:
A) Bekenstein's bound defines the maximum number of nats of information that can be contained in a spherical region as the circumference of that region divided by the reduced Compton wavelength associated with the total energy contained within that region,
B) the classical electron radius is equal to the fine structure constant times the reduced Compton wavelength of the electron.
Would you redo your calculation using the electron mass and the reduced Compton wavelength of the electron, you would obtain a value of $9.0647$ bits. However, you would obtain exactly the same value for a proton or whatever other elementary or composite particle you might chose. I would not attach any physical significance to these results.
Added: We currently don't have a consistent quantum gravity theory, and we don't even have an idea what would be the fundamental degrees of freedom in such a theory. Therefore any statement in response to questions like "how many bits/nats of information can be associated with an electron mass" runs the risk of leading to nonsense. Having said this, the holographic (Bekenstein-Hawking / black hole) bound seems more capable of providing reasonable leads. Using $4\pi$ times the square of the reduced Compton wavelength of the electron as area in the BH bound leads to an information content of $S/k = \pi \hbar c /G m^2$ nats. Here $m$ denotes the electron mass. This result for "the information content of a volume large enough to contain an electron" is in essence the square of the ratio of the Planck mass over the electron mass. That's a lot of nats.