 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,A coarser $100\mathrm{mV}$ spike estimate gives $6.2\times 10^6$ 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 $0.015\mathrm{mol}{\mathrm{L}}^{-1}\times 3\times 10^{-13}\mathrm{L}\times N_A\simeq 2.71\times 10^9$ 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 $1.8\times 10^6$ 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.