`R_("steel") = (L)/(KA) = (10^(-1)m)/(50(W//m-^(@)C)xx10^(-4)m^(2)) = (1000)/(50).^(@)C//W`. Similarly `R_(Cu) = (1000)/(400).^(@)C//W`

Junction `C` and `D` are identical in every respect and both will have same temperature. Consequently. The rod `CD` is in thermal equilibrium and no heat will flow through it. Hence it can be neglected in further analysis.

Now rod `BC` and `CE` are in series their equilvalet resistance is `R_(1) = R_(s) +R_(Cu)` similarly rods `BD` and `DE` are in series with same equilvalent resistance `R_(1) = R_(s) +R_(Cu)` these two are in parallel giving an equilvalent resistance of

`(R_(1))/(2) = (R_(S)+R_(Cu))/(2)`

This resistance is connected in series with rod `AB`. Hence the net equilvalent of combination is

`R = R_(steel) +(R_(1))/(2) (3R_(steel)+R_(Cu))/(2) = 500 ((3)/(50)+(1)/(400)).^(@)C//W`

Now, `i = (T_(H)-T_(C))/(R) =(125^(@)C)/(500((3)/(50)+(1)/(400)).^(@)C//W)=4`watt.