$Q=T . \Delta S$

According to $T-S$ diagram, the system absortbs heat $\mathrm{Q}_{1}$ from $\mathrm{A}$ to $\mathrm{B}$

$\mathrm{Q}_{1}=$ (Average temperature).$\Delta S$

$=\left(\frac{2 \mathrm{T}_{0}+\mathrm{T}_{0}}{2}\right)\left(2 \mathrm{S}_{0}-\mathrm{S}_{0}\right)$

$\mathrm{Q}_{1}=\frac{3}{2} \mathrm{T}_{0} \mathrm{S}_{0}$       $...(i)$

The system rejects heat $\mathrm{Q}_{2}$ from $\mathrm{B}$ to $\mathrm{C}$.

$\mathrm{Q}_{2}=\mathrm{T}_{0}\left(2 \mathrm{S}_{0}-\mathrm{S}_{0}\right)=\mathrm{T}_{0} \mathrm{S}_{0}$        $...(ii)$

From $\mathrm{C}$ to $\mathrm{A}$ entropy remains constant, hence The efficiency $\eta$ of the engine is therefore

$\eta=1-\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}=1-\frac{\mathrm{T}_{0} \mathrm{S}_{0}}{\frac{3}{2} \mathrm{T}_{0} \mathrm{S}_{0}}$

$=1-\frac{2}{3}=\frac{1}{3}$