Solution:

Given,
Thermal conductivity of brick wall, k = 1.2 W/mK
Thickness of brick wall, L = 0.15 m
Temperature of atmospheric air, T∞ = 30 °C
Outer surface temperature of wall, T1 = 100 °C
Emissivity of wall, ε = 0.8
Heat transfer coefficient, h = 20 W/m2K

Calculations:

1. Heat transfer rate through the wall:
Using Fourier's law of heat conduction, the rate of heat transfer through the wall is given by:
q = kA (T1 - T2) / L
where,
A = surface area of the wall
T2 = inner surface temperature of the wall

For a unit area of the wall, the surface area is given by:
A = L / 1 m

Therefore, the rate of heat transfer through the wall is given by:
q = k (T1 - T2) / L

2. Heat transfer rate from the outer surface of the wall:
The heat transfer rate from the outer surface of the wall to the surrounding air is given by:
q1 = hA (T1 - T∞)
where,
h = heat transfer coefficient

For a unit area of the wall, the heat transfer rate from the outer surface is given by:
q1 = h (T1 - T∞) / 1 m2

3. Heat transfer rate from the inner surface of the wall:
The heat transfer rate from the inner surface of the wall to the combustion gases is given by:
q2 = εσA (T2^4 - T∞^4)
where,
ε = emissivity of the wall
σ = Stefan-Boltzmann constant

For a unit area of the wall, the heat transfer rate from the inner surface is given by:
q2 = εσ (T2^4 - T∞^4) / 1 m2

4. Equating the heat transfer rates:
Since the wall is in steady state, the heat transfer rate through the wall, q, must be equal to the sum of the heat transfer rates from the outer and inner surfaces of the wall, q1 and q2, respectively:
q = q1 + q2

Substituting the values of q, q1, and q2, we get:
k (T1 - T2) / L = h (T1 - T∞) + εσ (T2^4 - T∞^4)

Solving for T2, we get:
T2 = { k / (hL) } (T1 - T∞) + [ εσ / (hL) ] (T1^4 - T∞^4)^(1/4)

Substituting the given values, we get:
T2 = { 1.2 / (20 × 0.15) } (100 - 30) + [ 0.8 × 5.67 × 10^-8 / (20 × 0.15) ] (100^4 - 30^4)^(1/4)

T2 = 498.6 K = 225.6 °C

Therefore, the inner surface temperature of the brick wall is 225.6 °C.