$Q = \frac{{KA\left( {{\theta _1} - {\theta _2}} \right)}}{l}$

Where, $K=coefficient\,of\,thermal\,conductivity$

$l=length\,of\,rod\,and\,A=Area\,of\,cross-section\,of\,rod$

If the junction temperature is $T$, then

${Q_{Copper}} = {Q_{Brass}} + {Q_{Steel}}$

$\frac{{0.92 \times 4\left( {100 - T} \right)}}{{46}} = \frac{{0.26 \times 4 \times \left( {T - 0} \right)}}{{13}} + $

$\frac{{0.12 \times4\times \left( {T - 0} \right)}}{{12}}$

$ \Rightarrow 200 - 2T = 2T + T$

$ \Rightarrow T = {40^ \circ }C$

$\therefore \,\,{Q_{Copper}} = \frac{{0.92 \times 4 \times 60}}{{46}} = 4.8\,cal/s$