= Solution
Let $s=[S_1]$ and treat the quencher concentration as a constant reservoir. Weak steady excitation, one emitting population, and the three specified processes give
$$
\frac{ds}{dt}=I_{\rm abs}-(k_F+k_Q[Q])s=0,
\qquad s=\frac{I_{\rm abs}}{k_F+k_Q[Q]}.
$$
Dividing the <fluorescence> photon-production rate by the absorbed <photon> rate therefore gives
$$
\boxed{\phi_Q=\frac{k_Fs}{I_{\rm abs}}=\frac{k_F}{k_F+k_Q[Q]}.}
$$
This assumes no saturation, photochemical loss, reabsorption, static association or additional excited-state decay, and that each counted absorbed <photon> reaches the stated $S_1$ population. In this restricted scheme $\phi_0=1$ and the unquenched lifetime is $\tau_0=1/k_F$.
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