Let temperature of junction be \(=\theta\)

Heat flowing to junction = heat out flowing

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

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

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

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

\(\frac{180}{11}=\theta\)

\(16.36^{\circ} C =\theta\)