You can basically treat the box as a whole as a body with an internal source equal to the sum of the power of the electronis (350 W), and then calculate what would the equilibrium be based on the convection radiation (and conductive if it rests on another surface) losses from the box.
I.e. for each of the faces calculate:
convective
$$P_{conv} = h \cdot A \cdot (T_{wall}-T_\infty)$$
where:
- $P_{conv}$: The power lost due to convection
- $h$ : the convection coefficient
- $A$ : is the surface area of the face of the box
- $T_{wall}$: the wall surface (this is the value you are after)
- $T_\infty$: is the ambient temperature
radiative:
The radiative losses are:
$$P_{rad} = \epsilon A \cdot \sigma \cdot T_{wall}^4$$
Where:
- $P_{rad}$ : the ratiative losses
- $\epsilon$: the emissivity of aluminium
- $\sigma$: Stefan-Boltzmann constant
- $A$ and $T_{wall}$ are the same as above
conductive:
(follows fouriers law), this should only be applied if a face is touching on another surface (ground).
$$P_{cond} = U \cdot A\cdot (T_{wall}- T_\infty)$$
where:
- U : Coefficinet of Heat transfer (units W/(m^2 K))
- $A, T_{wall}, T_\infty$: the same as above.
total losses
The total losses should be equal to the thermal power within. Therefore
$$P_{conv} + P_{cond} + P_{rad} = 350$$
You can solve the above equation (numerically usually), and obtain the $T_{wall}$ which can be thought of as the temperature within the box (it's slightly higher actually because of the conductivity of aluminium).
