= Solution
Use the supplied $-50\,\mathrm{kJ\,mol^{-1}}$ as a standard transformed <Gibbs free energy> of <ATP hydrolysis>, rather than replacing it with a different commonly quoted standard value. Approximate <thermodynamic activities> by <concentrations> divided by $c^\circ=1\,\mathrm M$, at an explicitly assumed $T=300\,\mathrm K$. The dimensionless reaction quotient is
$$
Q_h=\frac{([\mathrm{ADP}]/c^\circ)([P_i]/c^\circ)}{[\mathrm{ATP}]/c^\circ}
=\frac{(0.0008)(0.010)}{0.0016}=0.005.
$$
Consequently
$$
\Delta G_h=\Delta G_h^\circ+RT\ln Q_h
=-50.0-13.22=-63.22\,\mathrm{kJ\,mol^{-1}},
$$
and the reverse synthesis cost is $g_{\rm ATP}=63.22\,\mathrm{kJ\,mol^{-1}}$. Substituting bare millimolar numbers in this quotient without the <standard state> would give the wrong logarithm.
For outside-to-inside <proton> movement, the <membrane potential> is $\Delta\psi=-0.125\,\mathrm V$ and $\mathrm{pH}_{\rm in}-\mathrm{pH}_{\rm out}=0.5$, so $[H^+]_{\rm in}/[H^+]_{\rm out}=10^{-0.5}$. The molar <electrochemical potential> change is
$$
\Delta\widetilde\mu_H=F\Delta\psi+RT\ln(10^{-0.5})
=-F\left(0.125+\frac{2.303RT}{F}0.5\right).
$$
Thus the available <proton-motive force> magnitude is $0.15476\,\mathrm V$, supplying $14.93\,\mathrm{kJ\,mol^{-1}}$ per transported <proton>. Assuming one <proton> per c-subunit per revolution, three <ATP> per revolution and no leakage, slippage or transport overhead,
$$
\Delta G_{\rm turn}=3g_{\rm ATP}-n_cF\Delta p,\qquad
\boxed{n_c\geq\frac{3g_{\rm ATP}}{F\Delta p}=12.70.}
$$
The smallest integer permitting synthesis is therefore $\boxed{n_c=13}$ in this ideal budget. A ring of thirteen gives only a small driving margin; inefficiency or sustained finite output could require a larger ring. The result is an energetic lower bound and near-reversible estimate, not a unique structural determination from these data. Reasonable temperatures near $300$–$310\,\mathrm K$ leave the minimum integer thirteen. Additional <membrane transport> costs have been excluded explicitly.
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