We know that,
$W=P \Delta V$
Now, $W_{A B}=0$
$W_{B C}=3 P(2 V-V)=3 P V$
$W_{C D}=0$
$W_{D A}=-P V$
Total work done,
$W =W_{A B}+W_{B C}+W_{C D}+W_{D A}$
$=3 P V-P V=2 P V$
Heat given to the system from $A$ to $B$
$=n C_{V} \Delta T$
$=n \frac{3}{2} R \Delta T=\frac{3}{2} \times V \times \Delta P$
$=\frac{3}{2} \times V \times (3p-p)$
$=\frac{3}{2} \times V \times 2 P$
$=3 P V$
Similarly,
Heat given to the system from $B$ to $\bar{C}$
$=n C_{P} \Delta T$
$=n\left(\frac{5}{2} R\right) \wedge T$
$=\frac{5}{2}(3 P)(\Delta V)$
$=\frac{5}{2} \times(3 P) \cdot(2 V-V)$
$=\frac{5}{2} \times 3 P \times V$
$=\frac{15}{2} P V$
The heat is released from $C$ to $D$ and $D$ to $A$.
Efficiency $( \eta ) =\frac{\text { Work done by gas }}{\text { Heat given to gas }} \times 100$
$=\frac{2 P V}{3 P V+\frac{15}{2} P V} \times 100$
$=\frac{2 P V}{P V\left(3+\frac{15}{2}\right)} \times 100$
$=\frac{2}{\left(\frac{6+15}{2}\right)} \times 100$
$=\frac{2 \times 2}{21} \times 100$
$=\frac{4}{21} \times 100=19.04 \%$
