= Solution
The upper-state <density> is clamped near threshold once the <laser> operates, so excess pumping mainly increases <photon> <density> and output <power>. For approximately pump-independent absorption efficiency, $r=P_p/P_{p,\mathrm{th}}$, and both $P_{\mathrm{out}}$ and $f_{\mathrm{ro}}^2$ should rise approximately linearly with $P_p-P_{p,\mathrm{th}}$.
\b[The left ordinate is the square of the <frequency>], in $\mathrm{kHz}^2$, rather than <frequency> in kHz. Reading the original triangles gives a threshold near $12$–$13\ \mathrm{mW}$ and a slope about $1.0\ \mathrm{kHz}^2/\mathrm{mW}$. A representative straight-line fit to visually read marker centers gives $P_{p,\mathrm{th}}\simeq12.2\ \mathrm{mW}$ and $A=d f_{\mathrm{ro}}^2/dP_p\simeq1.02\times10^6\ \mathrm{Hz}^2/\mathrm{mW}$. Therefore
$$
A\simeq\frac1{4\pi^2\tau_c\tau_2P_{p,\mathrm{th}}},\qquad
\boxed{\tau_c\simeq\frac1{4\pi^2(0.013)(12.2)(1.02\times10^6)}
\simeq1.6\times10^{-7}\ \mathrm s.}
$$
The displayed precision is limited by graph reading and scatter; a lifetime around $0.15\ \mathrm{\mu s}$ is the meaningful result. The output squares have a compatible threshold and an approximately constant slope above it.
The round-trip time is
$$
t_{\mathrm{rt}}=\frac{2n_rL}{c}=\frac{2(1.46)(7)}c\simeq6.82\times10^{-8}\ \mathrm s.
$$
If mirror transmission is the only appreciable passive loss, <energy> decays by $R_1R_2=e^{-t_{\mathrm{rt}}/\tau_c}$ per round trip. For $R_1=1$,
$$
\boxed{R_2=e^{-t_{\mathrm{rt}}/\tau_c}\simeq0.65,}
$$
a <power> <reflectance> of roughly 65%. The exact exponential is preferable to replacing $1-R_2$ by $t_{\mathrm{rt}}/\tau_c$, since the inferred round-trip loss is not tiny. If distributed fibre loss is appreciable, the lifetime instead fixes the total round-trip retention; without a separate loss measurement, the bare mirror <reflectance> is not uniquely determined and is at least this retention factor.
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