The net work done in a PV diagram over a cycle is equal to the area enclosed by the cycle.
From the diagram: the path encloses a rectangle from 2 $\times$ 10$^5$ Pa to 4 $\times$ 10$^5$ Pa (pressure) and 2 m$^3$ to 6 m$^3$ (volume).
So, Area = $(4 - 2) \times 10^5 \cdot (6 - 2) = 2 \times 10^5 \cdot 4 = 8 \times 10^5$ J = 800 kJ
But as per options given in units of J, final answer = 800 J