= Solution
The original Panel A is readable: the baseline is roughly $-45\,\mathrm{mV}$ and the peak about $+40\,\mathrm{mV}$, giving a voltage excursion near $85\,\mathrm{mV}$. Using the specific <membrane capacitance> and surface area,
$$
C=0.01\times10^{-9}=10^{-11}\,\mathrm F,\quad
Q=C\Delta V\simeq8.5\times10^{-13}\,\mathrm C,
\quad\boxed{N_{\rm Na,net}=\frac{Q}{e}\simeq5.3\times10^6\text{ ions}.}
$$
A coarser $100\,\mathrm{mV}$ spike estimate gives $6.2\times10^6$ <ions>, illustrating the precision appropriate to the graph. These are the net monovalent charges needed to charge the <capacitor> if <sodium> supplies the inward charge; simultaneous opposing currents can make the actual <sodium> influx larger.
The cell contains $0.015\,\mathrm{mol\,L^{-1}}\times3\times10^{-13}\,\mathrm L\times N_A\simeq2.71\times10^9$ intracellular <sodium> <ions>. The capacitive estimate adds only about $0.20\%$, or $0.029\,\mathrm{mM}$. Thus a spike does not equilibrate bulk <concentrations>. The <sodium-potassium ATPase> maintains the <gradients> over many spikes, exporting three <sodium> <ions> per two <potassium> <ions> imported per <ATP>; the estimated <sodium> load alone corresponds to at least about $1.8\times10^6$ pump cycles. Rapid <repolarization> itself comes mainly from changing channel conductances, while sustained fast firing requires adequate energy supply and <gradient> restoration. The small stated volume and large area are used as supplied, without assuming a particular cell geometry.
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