 Solution

Including both flat ends of the cylindrical cell, its membrane surface area isThe capacitance is $C=(0.01\mathrm{F}{\mathrm{m}}^{-2})A=6.34602\mathrm{pF}$. The electric charge needed for the specified depolarization is thereforeIgnoring the ends gives $3.33\times 10^6$, only about one percent smaller. Inward movement of positive ions, or outward movement of negative ions, produces the stated positive change in the inside-minus-outside membrane potential.

The cell volume is $\pi r^2\ell =3.14159\times 10^{-16}{\mathrm{m}}^3$. Distributed over this volume, the transferred number would correspond to only $0.0177955\mathrm{mM}$. Relative to an illustrative, not source-specified, $100\mathrm{mM}$ internal ionic pool, that is a fraction $1.78\times 10^{-4}$. Thus a substantial change in membrane potential requires movement of only a tiny fraction of bulk ions; it does not require wholesale depletion of the intracellular solution. The separated electric charge is concentrated near the two membrane surfaces.