= Solution
{c}
Define the two concentration differences $u=C_a-C_b$ and $v=C_b-C_d$. For zero emissions and $R_a=R_b=R$, their <ordinary differential equation> is
$$
\frac d{dt}\begin{pmatrix}u\\v\end{pmatrix}
=\begin{pmatrix}-2r/R&s/R\\r/R&-s(1/R+1/R_d)\end{pmatrix}
\begin{pmatrix}u\\v\end{pmatrix}.
$$
Writing the positive decay rates as $\mu_i=1/\tau_i$, their characteristic equation is
$$
\mu^2-P\mu+Q=0,\qquad
P=\frac{2r+s}{R}+\frac s{R_d},\qquad
Q=\frac{rs(R_d+2R)}{R^2R_d}.
$$
Consequently
$$
\boxed{\tau_1=\frac2{P+\sqrt{P^2-4Q}},\qquad
\tau_2=\frac2{P-\sqrt{P^2-4Q}}.}
$$
For $r\gg s$, the first <normal mode> rapidly equalizes the fast reservoirs, giving $\mu_1\simeq2r/R$. The slower evolution treats them as a combined capacity $2R$ exchanging with $R_d$, giving $\mu_2\simeq s[1/(2R)+1/R_d]$. Therefore
$$
\boxed{\tau_1\simeq\frac R{2r},\qquad
\tau_2\simeq\frac{2RR_d}{s(R_d+2R)}\simeq\frac{2R}s\quad(R_d\gg R).}
$$
The timescale ratio is $\tau_2/\tau_1\simeq4r/s\gg1$ in the joint limit. Since the conserved <normal mode> has $\tau_3=\infty$, the hierarchy is $\tau_1\ll\tau_2\ll\tau_3$.
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