
Hence given figure can be redrawn as follows
( $\frac{Q}{t} = \frac{{({\theta _1} - {\theta _2})}}{R}$ and ${\left( {\frac{Q}{t}} \right)_{AB}} = {\left( {\frac{Q}{t}} \right)_{BC}}$
==> $\frac{{(90 - \theta )}}{{R/2}} = \frac{{(\theta - 0)}}{R}$
==> $180 - 2\theta = \theta $
==>$\theta = 60^\circ C$

$Reason :$ Peak emission wavelengths of a black body is proportional to the fourth-power of temperature.