= Solution
Including both end faces of the perfect cylinder, with $r=1\,\mu\mathrm m$ and $\ell=100\,\mu\mathrm m$,
$$
\mathcal A=2\pi r\ell+2\pi r^2=6.346\times10^{-10}\,\mathrm{m^2},\qquad
\mathcal V=\pi r^2\ell=3.142\times10^{-16}\,\mathrm{m^3}
=3.142\times10^{-13}\,\mathrm L.
$$
The <membrane capacitance> is $C=0.01\mathcal A=6.346\times10^{-12}\,\mathrm F$. The required net inward <electric charge> is $Q=C\Delta V=5.394\times10^{-13}\,\mathrm C$, or $Q/e\simeq3.37\times10^6$ monovalent ions. If <sodium> supplies that net <electric charge>, <Faraday constant> $F=N_Ae$ gives
$$
\boxed{\Delta[\mathrm{Na}^+]=\frac{C\Delta V}{F\mathcal V}
=1.780\times10^{-5}\,\mathrm{mol\,L^{-1}}
=0.0178\,\mathrm{mM}=17.8\,\mu\mathrm M.}
$$
The 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> $Q/\Delta t\simeq0.270\,\mathrm{nA}$; 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.
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