A heat source at T=103K is connected to another heat reservoir at T=102K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is 0.1 KW/m, the energy flux through it in the steady state is

(dQ/dt)=kAΔT/l ⇒ (dQ/dt)=(0.1)(1)(103-102)/1 = 10 W/m².
Assuming the area of the copper slab is 1m², then the energy flux is 10W/m² x 1m² = 10 W. There must be an error in either the question or the provided solution. The correct calculation is shown above. The given solution appears to be using a value of 900 for the area. The options provided do not include this value.