Acylindrical log 18 cm in diameter and 75 cm long is glowing red hot in a fireplace. if it's emitting radiation at the rate of 38 kw , what is its temperature? the log's emissivity is essentially 1

To calculate the temperature of the log we need the Stefan-Boltzmann's law:

P=A*ε*σ*T⁴, where P is the power emitted by the body,
A is the total surface area of the body, in our case it is a cylinder so A=r²π*h where r is the radius of the base of the cylinder and h is the height of the cylinder, 
σ is the Stefan-Boltzmann constant, T is temperature and ε is emissivity .
Here we are approximating the log to be a black body. 

The area of the cylinder:

A=r²*π*h, r=d/2=0.75/2=0.375 m, where d is the diameter, h=0.18 m

A=0.07948 m²

Lets solve the equation for temperature T:

T⁴=P/(σ*ε*A) and take the 4th root to get T:

T=⁴√{P/(σ*ε*A)}=⁴√{38000/((5.67*10^-8)*1*0.07948)} = ⁴√(8.432*10^12)= 1704.06 C

So the temperature of the log is T= 1704 C

The answer is d) by requesting advice on science internet sites