Answer

For the adiabatic strokes, $Q=0$ and $W=c_V(T_f-T_i)$. (See a previous example.)

For the isochoric (constant volume) paths, $W=0$ and $Q=c_V(T_f-T_i)$.

Also

$$T_2=\left({V_1\over V_2}\right)^{\gamma-1} T_1 \qquad\hbox{and}\qquad T_3=\left({V_1\over V_2}\right)^{\gamma-1} T_4$$

The total work done by the system is

$$W= -(W_{12}+W_{34}) =c_V(T_1-T_2+T_3-T_4)$$

so

$$\eta={W\over Q_H}={T_1-T_2+T_3-T_4\over T_3-T_2}=1-\left({V_2\over V_1}\right)^{\gamma-1}$$