Ts Tw
| R | 2/3
600 -- -- --*---www-------www--- 70
I |
| 1/3
---www--- 70

If the stove temp is 600 F, I = 0.1714x10^-8(600+460)^4 = 2164 Btu/h-ft^2.
A 90% reflective shield would absorb about 10% of this, ie 216 Btu/h-ft^2.
The linearized radiation conductance R = 1/(4x0.1714x10^-8(70+460)^3x0.1)
= R9.8, with 2/3 airfilm conductances. Here's one equivalent circuit:

I ------------------------
Ts Tw
1/3 | 9.8 | 2/3
---www---*---www-------www--- 70 F
|
| 142 F = 70+216x1/3
---
_
|
-

What are the shield and wall temps Ts and Tw?

It's just Ohm's law, with different units:

I = (142-70)/(1/3+9.8+2/3) = 6.7 Btu/h, so Tw = 70 + 6.7x2/3 = 74.4 F,
and Ts = 74.4 + 6.7x9.8 = 139.7.

Nick