= Solution
Use dimensional quantities $T_d,t_d$ and set $T_d=T_0T$, $t_d=t_0t$. Dividing the <energy balance> by $\sigma\gamma T_0^4$ and the <albedo> equation by its time scale gives
$$
\boxed{\varepsilon T_t=2(1-a)-T^4,\quad a_t=a^*(T)-a,\quad a^*(T)=\frac{2-\tanh[\mu(T^4-1)]}{4},\quad \varepsilon=\frac{\rho c_pd}{t_0\sigma\gamma T_0^3}=\frac{8\rho c_pdT_0}{Qt_0}.}
$$
Thus $\varepsilon$ is the ratio of the atmospheric thermal response time to the much longer <albedo> response time. Converting kilometres to metres and years to seconds gives $C=10^7\ \mathrm{J\,m^{-2}K^{-1}}$, $T_0\simeq266.3\ \mathrm K$, $t_0\simeq3.156\times10^{10}\ \mathrm s$, and
$$
\boxed{\varepsilon\simeq4.93\times10^{-4}.}
$$
The thermal time is about half a year, compared with a millennium for <albedo> adjustment. This is a <singular perturbation>: setting $\varepsilon=0$ removes the independent <initial condition> for <temperature>.
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