$\text{A} \to \text{B}$ isochoric as $\mathrm{P}$ is directly proportional to $\mathrm{T}$

${\text{W}}_{\text{AB}}=0$

${\text{Q}}_{\text{AB}}=\text{n}{\text{C}}_{\text{v}}\left(3{\text{T}}_{}-{\text{T}}_{}\right)$

       $=\left(\frac{5}{2}\right)\text{nR}2{\text{T}}_{}=5{\text{nRT}}_{}$

$\mathrm{B} \to \mathrm{C}$isothermal and thus $\mathrm{Q} = \mathrm{W}$ as there won't be any change in internal energy.
${\text{W}}_{\text{BC}}=\text{nR}3{\text{T}}_{}ln\left(\frac{{\text{P}}_{\text{B}}}{{\text{P}}_{\text{C}}}\right)=3{\text{nRT}}_{}\text{ln 3}$

       $=3{\text{nRT}}_{}\text{ln}3={\text{Q}}_{\text{BC}}$

$\mathrm{C} \to \mathrm{A}$isobaric process
${\text{W}}_{\text{CA}}=\text{P}\left({\text{V}}_{\text{A}}-{\text{V}}_{\text{C}}\right)=\text{nR}\left({\text{T}}_{\text{A}}-3{\text{T}}_{}\right)=-2\text{n}{\text{RT}}_{}$

Efficiency of a cyclic process is given by

                  $\eta =\frac{\text{Work done in cyclic process}}{\text{Heat supplied to the system}}=\frac{\text{W}}{{\text{Q}}_{\text{in}}}$

Efficiency $\eta =\frac{3{\text{nRT}}_{\text{A}}ln3-2{\text{nRT}}_{\text{A}}}{5{\text{nRT}}_{\text{A}}+3{\text{nRT}}_{\text{A}}ln3}$

                  $=\frac{\text{3 ln 3}-2}{5+3\text{ln 3}}$