= Solution
Let $r=a_{\rm ADP}/a_{\rm ATP}$ and use a one-molar <standard state>, so the <inorganic phosphate> activity is approximately $a_P=0.006$. With the source's stated standard <hydrolysis> value,
$$
\Delta G_{\rm hyd}=-50{,}000+RT\ln(0.006r)\quad\mathrm{J\,mol^{-1}}.
$$
The downhill proton-transfer <Gibbs free energy> is $F\Delta p$ for the signed driving <force> $\Delta p=-0.150$ V. Ideal reversible balance of one complete turn requires
$$
-3\Delta G_{\rm hyd}+10F\Delta p=0,
\qquad \Delta G_{\rm hyd}=\frac{10}{3}F\Delta p=-48.243\ \mathrm{kJ\,mol^{-1}}.
$$
Solving yields
$$
\boxed{r=\frac1{0.006}\exp\!\left[\frac{50{,}000+(10/3)F\Delta p}{RT}\right]
\simeq3.29\times10^2\quad(T=310.15\ \mathrm K).}
$$
At 298.15 K the estimate is $3.39\times10^2$. <Temperature> is not specified in this paragraph; the first value explicitly adopts physiological <temperature> rather than pretending it is supplied. The unexpectedly large <ADP>/<ATP> ratio follows from using the \b[printed standard] value $-50$ kJ/mol with the stated dilute <inorganic phosphate>. One must not silently replace that value by another convention or treat it as the already concentration-corrected cellular <hydrolysis> <Gibbs free energy>.
This estimate assumes ideal <thermodynamic activities>, consistent <pH> and <magnesium> conventions, tight coupling of ten <protons> to three <ATP>, no leak or transport overhead, and operation close to reversible balance. A working cell is not in global equilibrium. Net synthesis at this fixed driving <force> requires $\Delta G_{\rm turn}<0$, hence $r$ \b[above] this reversible-balance threshold; finite <chemical kinetics> and dissipation prevent the <stoichiometry> alone from determining an exact physiological ratio.
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