Answer
See the detailed answer below.
Work Step by Step
First, we need to find $P_1$ at the initial point using the ideal gas law.
$$P_1=\dfrac{nRT_1}{V_1}=\dfrac{(0.20)(8.31)(600)}{(2000\times 10^{-6})}=\bf 498.6\;\rm kPa$$
Now we need to find $P_2$; noting that from 1 to 2 is an isothermal process.
$$P_2=\dfrac{nRT_1}{V_2}=\dfrac{(0.20)(8.31)(600)}{(4000\times 10^{-6})}=\bf 249.3\;\rm kPa$$
Now we need to find $P_3$; where from 2 to 3 the gas undergoes an isochoric process, so $V_2=V_3$
$$P_3=\dfrac{nRT_3}{V_2}=\dfrac{(0.20)(8.31)(300)}{(4000\times 10^{-6})}=\bf 124.65\;\rm kPa$$
Now we need to find $P_4$; where from 3 to 4 the gas undergoes an isothermal process, so $T_3=T_4$,
$$P_4=\dfrac{nRT_3}{V_2}=\dfrac{(0.20)(8.31)(300)}{(2000\times 10^{-6})}=\bf 249.3\;\rm kPa$$
See the graph below;
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Now we need to find the work done by the system in one cycle.
$$W_s=(W_s)_{1\rightarrow2}+(W_s)_{2\rightarrow3}+(W_s)_{3\rightarrow4}+(W_s)_{4\rightarrow1} $$
The two process from 2 to 3 and from 4 to 1 are isochoric processes, so the work done here is zero.
$$W_s=(W_s)_{1\rightarrow2}+0+(W_s)_{3\rightarrow4}+0$$
The two other processes are isothermals, so
$$W_s=nRT_1\int_1^2\dfrac{dV}{V} +nRT_3\int_3^4\dfrac{dV}{V} $$
$$W_s=nRT_1\ln\left[\dfrac{V_2}{V_1} \right]+nRT_3\ln\left[\dfrac{V_4}{V_3} \right]$$
Plugging the known;
$$W_s=(0.20)(8.31)\left((600)\ln\left[\dfrac{4000}{2000} \right]+ (300)\ln\left[\dfrac{2000}{4000} \right]\right)$$
$$W_s=\color{red}{\bf 345.6}\;\rm J$$
To find the engine efficiency, we need to find $Q_{in}$ so we have to find the heat exchange in each process.
During the two isothermal processes, the thermal energy does not change, so
$$Q=W_s$$
Hence,
$$Q_{1\rightarrow2}=(W_s)_{1\rightarrow2}=nRT_1\ln\left[\dfrac{V_2}{V_1} \right]$$
$$Q_{1\rightarrow2}=(0.20)(8.31) (600)\ln\left[\dfrac{4000}{2000} \right]=\bf 691.2\;\rm J$$
$$Q_{3\rightarrow4}=(W_s)_{3\rightarrow4}=nRT_3\ln\left[\dfrac{V_4}{V_3} \right]$$
$$Q_{3\rightarrow4}=(0.20)(8.31) (300)\ln\left[\dfrac{2000}{4000} \right]=\bf -345.6\;\rm J$$
In the two isochoric processes, the work done is zero, so
$$Q=\Delta E_{th}$$
Hence,
$$Q_{2\rightarrow3}=(\Delta E_{th})_{2\rightarrow3}=\frac{3}{2}nR(T_3-T_2)$$
$$Q_{2\rightarrow3}=\frac{3}{2}(0.20)(8.31) (300-600)=\bf -747.9\;\rm J$$
$$Q_{4\rightarrow1}=(\Delta E_{th})_{4\rightarrow1}=\frac{3}{2}nR(T_1-T_4)$$
$$Q_{4\rightarrow1}=\frac{3}{2}(0.20)(8.31) (600-300 )=\bf 747.9\;\rm J$$
Thus,
$$Q_{in}=Q_{1\rightarrow2}+Q_{4\rightarrow1}=691.2+747.9$$
$$Q_{in}=\color{blue}{\bf 1439.1}\;\rm J$$
Therefore,
$$\eta=\dfrac{W_s}{Q_{in}}=\dfrac{345.6}{1439.1}=\color{red}{\bf0.24}$$
