Electron Electric Field Mass?

I am confused of whether or not the expected electromagnetic field generated by the point-like electric charge of the electron distributed smoothly across space as a probability distribution creates the presence of an effective field mass.

I know that the probability current of the electron is $\langle \gamma^{0k} \rangle$, which is conserved. Multiplying the probability current of the electron by its total charge $q$ gives the charge current density $J^k = q\langle \gamma^{0k} \rangle$. From the current I can calculate the expected four-potential field as \begin{align} A^k(r,t) = \int \dfrac{J^k(r',t-c/r')}{|r-r'|} d^3 r' . \end{align}

From $A^k$ I can calculate the expected electromagnetic fields as $F^{jk} = \partial ^j A^k - \partial^k A^j$. Finally I can calculate the expected electromagnetic energy as \begin{align} U_{\text{eff}} = \dfrac{1}{8\pi} \int F^{jk}F^{jk} d^3 r . \end{align} Applying Einstein's relation energy is proportional to mass, I get the following effect field mass of the electron as \begin{align} M_{\text{eff}} = \dfrac{U_{\text{eff}}}{c^2} . \end{align}

Is $M_{\text{eff}}$ a real observable? I was unable to find an analytic solution for $M_{\text{eff}}$, however I did compute $M_{\text{eff}}$ for Gaussian distributed probability functions for the electron with varying standard deviation (spacial localization), and I noticed that for standard deviations of the order of $10^{-10}$ meters (lattice size), $M_{\text{eff}}$ was very small in comparison to the electron's rest mass. When the standard deviation was $10^{-15}$ meters (nucleus size), the $M_{\text{eff}}$ was comparable in size to the rest mass of the electron.

If the electron is point-like, the Poynting energy expression is infinite and thus it is useless for calculation of the energy of the electron in the Einstein sense $E=mc^2$, which is finite.
According to common interpretation of some fine experiments, electrons are known to be smaller than $10^{-18}~$m today and are possibly points.