I think that first we should be more explicit about the analogy. We know what the circuit represents normally, which is voltage and current. For this system, we should correlate the analog to real values as follows (hat tip to Ilmari's comment).
- Voltage - Temperature (K)
- Current - Heat flow (W)
- Capacitor charge - Heat content (J)
I see why the sun would be modeled as a current source and not a voltage source, because the sun's cycle represents changes in emissivity and not temperature (and by all means let me know if I'm wrong). The number of photons changes, but not their frequency.
But here's where I take issue: if the sun is a current source, the resistor after the sun doesn't make sense to me. I'll show (in a bit here) how to relate the temperature of the Earth to the energy flow into space, but the temperature of the sun doesn't change! So what are we relating by proposing a resistor between the sun and the Earth? If the absorptivity of the Earth changed that would affect things, but the proposition of a resistor is to say that a resistance to flow exists proportional to the flow, and I simply see no such thing in the sun-Earth connection. So I would just take that resistor out.
Again, you're looking for
- Resistivity of sun to Earth $R_{earth}$
- Resistivity of Earth to space $R_{loss}$
- Capacitance of Earth $C_{Earth}$
The main thing we need to do is to create a linearized formulation. I think I have an appropriate example in mind for $R_{loss}$, from climate science. Fundamentally, they write:
$$\Delta T_f = \lambda \Delta F$$
From Wikipedia, with $\Delta T_f$ as the temperature change due to a change in surface heat flow (all the relevance in the world to your question) of $\Delta F$. The parameter $\lambda$ comes from the derivative of the blackbody radiation correlation of the type $\sigma T^4$, but we need not concern ourselves with this.
(aside: my background researching this comes from a question or two of my own on the subject on Physics SE site, although that was concerned with $\Delta F$, $\lambda$ is comparatively easier I think)
For the purposes of finding $R_{loss}$, we know $\lambda =0.8 \frac{K}{\frac{W}{m^2}}$, and this is specific to Earth. The sun would have its own corresponding value (which is easy to calculate), but again, I'm not in agreement that that's relevant to the problem. Now I need to relate this to your circuit. Well, like I said, temperature is the voltage and the flow ($\Delta F$) is proportional to the current. You might prefer to multiply the units by the area of the Earth to get rid of the $m^2$ units. To the extent that you've adjusted the units correctly, I can just go ahead and assert that $\lambda = R_{loss}$.
For the specific heat of the Earth, I'll write a dumb equation in order to have something.
$$C_{earth} = \sum_i Cp_i m_i $$
Where $i$ represents every component on Earth. This includes the atmosphere, ocean, ground, and whatever else. Note that $C$ is not the same units as $Cp$, by a mass unit. I think it comes out to $J/k$, which is like change in charge over change in voltage. Capacitance, sure.
The remaining problem for $C$ is that of inclusion. Many thermal masses on Earth won't be relevant for the time frame of the thermal cycles (like the mantle). Good luck figuring out where to draw the line on that one. And be mindful of your units!