The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is:

Efficiency $\eta =\frac{\text{Work done}}{\text{Heat absorbed}}$.In a $T-S$ diagram, Work done $W=$ Area enclosed by the cycle.
$$W=\text{Area of triangle}=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times (2S_0-S_0)\times (2T_0-T_0)=\frac{1}{2}{S}_0{T}_0$$
Heat absorbed $Q_{in}$ occurs during processes where entropy increases. Here, it is along the path from $(S_0,T_0)$ to $(2S_0,2T_0)$ and the vertical path.Actually, let's look at the paths:$A(S_0,T_0)$ to $B(2S_0,2T_0)$ (Heating and expansion)$B(2S_0,2T_0)$ to $C(2S_0,T_0)$ (Isentropic cooling - no heat)$C(2S_0,T_0)$ to $A(S_0,T_0)$ (Isothermal compression - heat rejected)Total heat in $={\int }_A^{B}TdS=\text{Area under AB}=\text{Area of rectangle}+\text{Area of triangle}$.
$$Q_{in}=S_0{T}_0+\frac{1}{2}{S}_0{T}_0=\frac{3}{2}{S}_0{T}_0$$
$$\eta =\frac{W}{Q_{in}}=\frac{0.5S_0{T}_0}{1.5S_0{T}_0}=\frac{1}{3}$$