`(b)` Let `theta^(@)C` be the temperature at `B`. Let `Q` is the heat flowing per second from `A` to `B` on account of temperature difference by conductivity.
`:. Q=(KA(90-theta))/(l)`…….`(i)`
Where `K=` Thermal conducitivity of the rod
`A=` Area of cross-section of the rod
`l=` Length of the rod
By symmetry, the same will be the case for heat flow from `C` to `B`
`:.` The heat flowing per second from `B` to `D` will be
`2Q=(KA(theta-0))/(l)`......`(ii)`
Diveding equation `(ii)` by equation `(i)`
`2=(theta)/(92-theta)`
`implies theta=60^(@)`