Answer :
To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzman law that is responsible for calculating radioactive energy.
Mathematically this expression can be given as
[tex]P = \sigma Ae\Delta T^4[/tex]
Where
A = Surface area of the Object
[tex]\sigma =[/tex] Stefan-Boltzmann constant
e = Emissivity
T = Temperature (Kelvin)
Our values are given as
[tex]A = 1.36m^2[/tex]
[tex]\Delta T^4 = T_2^4 -T_1^4 = 307^4-T_1^4[/tex]
[tex]\sigma = 5.67*10^{-8} J/(s m^2 K^4)[/tex]
[tex]P = 122J/s[/tex]
[tex]e = 0.7[/tex]
Replacing at our equation and solving to find the temperature 1 we have,
[tex]P = \sigma Ae\Delta T^4[/tex]
[tex]P = \sigma Ae (T_2^4 -T_1^4)[/tex]
[tex]122 = (5.67*10^{-8})(1.36)(0.7)(307^4-T_1^4)[/tex]
[tex]T_1 = 285.272K = 12.122\°C[/tex]
Therefore the the temperature of the coldest room in which this person could stand and not experience a drop in body temperature is 12°C