Let temperature of junction be $=\theta$

Heat flowing to junction = heat out flowing

$\frac{K A}{30}(30-\theta)=\frac{K A}{20}(\theta-20)+\frac{K A}{10}(\theta-10)$

$\frac{(30-\theta)}{3}=\frac{(\theta-20)}{2}+\frac{(\theta-10)}{1}$

$\frac{(30-\theta)}{3}=\frac{\theta-20+2 \theta-20}{2}$

$\frac{60-2 \theta}{3}=9 \theta-120$

$\frac{180}{11}=\theta$

$16.36^{\circ} C =\theta$