 Solution

Including both end faces of the perfect cylinder, with $r=1\mu \mathrm{m}$ and $\ell =100\mu \mathrm{m}$,The membrane capacitance is $C=0.01\mathcal{A}=6.346\times 10^{-12}\mathrm{F}$. The required net inward electric charge is $Q=C\Delta V=5.394\times 10^{-13}\mathrm{C}$, or $Q/e\simeq 3.37\times 10^6$ monovalent ions. If sodium supplies that net electric charge, Faraday constant $F=N_{A}e$ givesThe final concentration is about $15.0178\mathrm{mM}$, only $0.119\%$ above the starting value. Neglecting the end faces changes the estimate by $1\%$, giving about $17.6\mu \mathrm{M}$. Over $2\mathrm{ms}$ the capacitive electric charge corresponds to mean net electric current $Q/\Delta t\simeq 0.270\mathrm{nA}$; the duration does not otherwise enter the total electric charge.

This is a net capacitive estimate, not a measurement of total sodium influx. Simultaneous outward potassium electric current makes actual sodium entry larger, and most bulk ions remain near their original concentrations during one spike. Long-term pumps restore gradients after repeated firing; they are not required to replace the entire intracellular sodium pool during the two-millisecond voltage rise.