One mole of a monatomic ideal gas undergoes a cyclic process A→ B→ C→ A. Process A→ B is isochoric, raising the pressure from P0 to 3P0 at volume V0.…

Heat is absorbed in $A\to B$ and $B\to C$; it is rejected in $C\to A$. For $A\to B$: $Q_{AB}=C_V\Delta T=\tfrac32\,\Delta(PV)=\tfrac32(3P_0V_0-P_0V_0)=3P_0V_0$. For $B\to C$ (isothermal, $T_B=T_C$): $P_CV_C=3P_0V_0$ so $V_C=3V_0$ and $Q_{BC}=W_{BC}=3P_0V_0\ln 3$. In $C\to A$ the work done by the gas is $P_0(V_0-3V_0)=-2P_0V_0$. Net work $W=3P_0V_0\ln3-2P_0V_0=1.2958\,P_0V_0$. Heat absorbed $Q_{in}=3P_0V_0+3P_0V_0\ln 3=6.2958\,P_0V_0$. Hence $\eta=1.2958/6.2958=0.2058\approx 20.6\%$. Because the cycle contains an isothermal arc, $\Delta U=0$ over $B\to C$, so all heat absorbed there becomes work; the isobaric leg is the only place heat leaves the gas. The value $52.3\%$ ignores the compression work, $39.3\%$ divides by only the isothermal heat, and $43.2\%$ divides by $3P_0V_0$ alone.