The work done during the isobaric processes $2-3$ and $1-4$ are as follows:

$\mathrm{W}_{2-3}=\mathrm{P}_{2}\left(\mathrm{V}_{3}-\mathrm{V}_{2}\right)$

$w_{1-4}=P_{1}\left(V_{1}-v_{4}\right)$

Total work done $=\mathrm{P}_{2}\left(\mathrm{V}_{3}-\mathrm{V}_{2}\right)+\mathrm{P}_{1}\left(\mathrm{V}_{1}-\mathrm{V}_{4}\right)$

$\mathrm{W}_{\mathrm{T}}=\mathrm{P}_{2} \mathrm{V}_{3}-\mathrm{P}_{2} \mathrm{V}_{2}+\mathrm{P}_{1} \mathrm{V}_{1}-\mathrm{P}_{1} \mathrm{V}_{4}$

Three moles has been given, so

$\mathbf{P V}=n R T=3 R T$

$\mathrm{W}_{\mathrm{T}}=3 \mathrm{RT}_{3}-3 \mathrm{RT}_{2}+3 \mathrm{RT}_{1}-3 \mathrm{RT}_{4}$

$=3 \mathrm{R}\left[\mathrm{T}_{1}+\mathrm{T}_{3}-\mathrm{T}_{2}-\mathrm{T}_{4}\right]$

$=3 \mathrm{R}[400+2400-800-1200]$

$=3 \mathrm{R} \times 800=20 \times 10^{3} \mathrm{J}=20 \mathrm{kJ}$