Treating the given network of rods in terms of thermal resistance R_x and R_y with

${\text{R}}_{\text{x}}=\frac{\text{L}}{\text{A}\times \text{0.92}}\text{and}{\text{R}}_{\text{y}}=\frac{\text{L}}{\text{A}\times \text{0.46}}\left[\text{as R}=\frac{\text{L}}{\text{AK}}\right]$

so that if   R_x = R,  R_y = 2R_x = 2R

Now as in this bridge [(P/Q) = (R/S)], so the bridge is balanced, i.e., the temperature at junctions C and D is equal and the rod CD becomes ineffective as no heat will flow through it.

Now as the thermal resistance of the bridge between junctions B and E is

$\frac{1}{{\text{R}}_{\text{BE}}}=\frac{1}{\left(\text{R}+\text{R}\right)}+\frac{1}{\left(\text{2R}+\text{2R}\right)}\text{,}\text{i.e.,}{\text{R}}_{\text{BE}}=\frac{4}{3}\text{R}$

The total resistance of bridge between A and E will be

      R_eq = R_AB + R_BE = 2R + (4/3) R = (10/3) R

So the net rate of flow of heat through the bridge will be

$\frac{\text{dQ}}{\text{dt}}=\frac{\text{Δ}\text{θ}}{{\text{R}}_{\text{eq}}}=\frac{\left(\text{60}-\text{10}\right)}{\left(\text{10/3}\right)\text{R}}=\frac{\text{15}}{\text{R}}$

Now if T_B is the temperature at B,

$[\frac{\text{dQ}}{\text{dt}}{]}_{\text{AB}}=\frac{\text{Δ}\text{θ}}{{\text{R}}_{\text{AB}}}=\frac{\text{60}-{\text{T}}_{\text{B}}}{2\text{R}}$

$\text{But}[\frac{\text{dQ}}{\text{dt}}{]}_{\text{AB}}=\frac{\text{dQ}}{\text{dt}}\text{,}\text{i.e.,}\frac{\text{60}-{\text{T}}_{\text{B}}}{2\text{R}}=\frac{\text{15}}{\text{R}}\text{,}\text{i.e.,}{\text{T}}_{\text{B}}={\text{30}}^{\text{o}}\text{C}$

Also at B

$[\frac{\text{dQ}}{\text{dt}}{]}_{\text{AB}}=[\frac{\text{dQ}}{\text{dt}}{]}_{\text{BC}}+[\frac{\text{dQ}}{\text{dt}}{]}_{\text{BD}}$

$\text{i.e.,}\frac{\text{15}}{\text{R}}=\frac{\text{30}-{\text{T}}_{\text{C}}}{\text{R}}+\frac{\text{30}-{\text{T}}_{\text{D}}}{\text{2R}}$

and as     T_C = T_D = T,    30 = 3 (30 - T),

i.e.,          T_C = T_D = T = 20^o C