Let total resistance of the ring be $R$

for $\theta ={180}^o$ , we can consider the ring as a parallel combination of two sections of resistance $\frac{R}{2}$ each.

$\therefore R_{eq}=\frac{R}{4}$

$\therefore$ Rate of heat flow

$P_1=1.2=\frac{\Delta T}{\frac{R}{4}}$

$\Rightarrow ∆T=0.3R$

for, $\theta ={90}^o$ we can consider the ring as a parallel combination of two sections of resistance, $\frac{R}{4}$ & $\frac{3R}{4}$

$\therefore R_{eq}=\frac{\frac{R}{4}.\frac{3R}{4}}{\frac{R}{4}+\frac{3R}{4}}=\frac{3R}{16}$

$\therefore P_2=\frac{\Delta T}{\frac{3R}{16}}=\frac{16}{3R}\left(0.3R\right)=1.6 \mathrm{w}\mathrm{a}\mathrm{t}\mathrm{t}$