Work done $=Q_{\text {in }}-Q_{\text {out }}=300-240=60 {J}$
Efficiency $=\frac{W}{{Q}_{\text {in }}}=\frac{60}{300}=\frac{1}{5}$
efficiency $=1-\frac{{T}_{2}}{{T}_{1}}$
$\frac{1}{5}=1-\frac{400}{{T}_{1}} \Rightarrow \frac{400}{{T}_{1}}=\frac{4}{5}$
${T}_{1}=500\, {k}$
Step $1$ It is first compressed adiabatically from volume $8.0 \,m ^{3}$ to $1.0 \,m ^{3}$.
Step $2$ Then expanded isothermally at temperature $T_{1}$ to volume $10.0 \,m ^{3}$.
Step $3$ Then expanded adiabatically to volume $80.0 \,m ^{3}$.
Step $4$ Then compressed isothermally at temperature $T_{2}$ to volume $8.0 \,m ^{3}$.
Then, $T_{1} / T_{2}$ is

