One mole of a monatomic ideal gas undergoes the cycle A→ B→ C→ A in the P–V plane: A(V0,P0)→ B(V0,2P0) at constant volume, B→ C(2V0,2P0) at constant…

Net work is the area of triangle $ABC$: $\frac12 V_0 P_0$. Heat in $AB$ (isochoric): $\Delta U=\frac32\Delta(PV)=\frac32P_0V_0$. Heat in $BC$ (isobaric): $Q=\frac52\Delta(PV)=\frac52(4P_0V_0-2P_0V_0)=5P_0V_0$. On $CA$ the line is $P=(P_0/V_0)V$, so $T\propto V^2$ and $dQ=dU+P\,dV=3aV\,dV+aV\,dV=4aV\,dV$ with $a=P_0/V_0$. The gas is compressed along $CA$, so $dQ<0$ throughout: heat is rejected there and does not count as input. Total heat absorbed is $\frac32+5=\frac{13}{2}P_0V_0$, hence $\eta=\frac{1/2}{13/2}=\frac1{13}$. Including $CA$ wrongly or using the wrong $C_p$ gives the other values. Always check the sign of $dQ$ along a curved or inclined process before adding heats, because only the heat absorbed counts in the denominator of the efficiency of a cycle, and the cycle's net work equals the net heat exchanged.