Gas is monoatomic.Hence $C_{v}=\frac{3}{2} R, C_{p}=\frac{5}{2} R$

Heat absorbed during $\mathrm{AB}, \Delta Q_{A B}=n C_{v} \Delta T=\frac{3}{2} n R \Delta T=\frac{3}{2} V \Delta P=3 P_{o} V_{o}$

Heat absorbed during $\mathrm{BC}, \Delta Q_{B C}=n C_{p} \Delta T=\frac{5}{2} n R \Delta T=\frac{5}{2} P \Delta V=\frac{15}{2} P_{o} V_{o}$

Hence, net heat absorbed $=\Delta Q=\Delta Q_{A B}+\Delta Q_{B C}=\frac{21}{2} P_{o} V_{o}$

Also, net work done $=\Delta W$ =area under $P-V$ graph $=2 P_{o} V_{o}$

Hence efficiency $=\frac{\Delta W}{\Delta Q}=\frac{4}{21}$