Answer:
Q=116.37 W
Step-by-step explanation:
Given that
Emissivity ,ε= 0.200
Surface temperature ,T₁ = 10⁰ C = 283 K
Surrounding T₂ = - 15⁰ C = 258 K
Area ,A= 1.6 m²
The net heat transfer given as

σ = 5.67 x 10⁻⁸
The temperature should be in Kelvin.
Now by putting the values

Q=116.37 W
Therefore answer will be 116.37 W.