One mole of a monatomic ideal gas undergoes a cycle represented on a P-V diagram by a rectangle: (P0,V0)→(2P0,V0)→(2P0,2V0)→(P0,2V0)→(P0,V0), all…

Work per cycle $W=(2P_0-P_0)(2V_0-V_0)=P_0V_0$. Heat is absorbed only on the isochoric rise $A\to B$ and the isobaric expansion $B\to C$ (on the other two legs temperature falls, so heat is rejected). For $\gamma$ of a monatomic gas, $C_V=\tfrac32R$, $C_P=\tfrac52R$. Using $PV=RT$: $Q_{AB}=\tfrac32(2P_0V_0-P_0V_0)=\tfrac32P_0V_0$ and $Q_{BC}=\tfrac52(4P_0V_0-2P_0V_0)=5P_0V_0$. $Q_{in}=\tfrac{13}{2}P_0V_0$, so $\eta=\tfrac{2}{13}$. The extreme temperatures are $T_{\min}=P_0V_0/R$ (at $A$) and $T_{\max}=4P_0V_0/R$ (at $C$), giving $\eta_{\rm Carnot}=1-\tfrac14=\tfrac34$. Ratio $=\tfrac{2/13}{3/4}=\tfrac{8}{39}$. The option $2/13$ is $\eta$ itself (forgetting to divide by $\eta_{\rm Carnot}$); $4/13$ and $1/3$ come from mis-computing the heat input or the Carnot efficiency.