To solve this problem, we need to calculate the work done by the system in the cyclic process ABCDA.$\\$The work done in a cyclic process is equal to the area enclosed by the cycle on the PV diagram.$\\$Let's analyze the given PV diagram:$\\$• The cycle is a rectangle with vertices A, B, C, and D.$\\$• The coordinates of these points are:$\\$ A: $(V_0, P_0) \\$ B: $(2V_0, 2P_0) \\$ C: $(V_0, 3P_0) \\$ D: $(2V_0, P_0) \\$The area of the rectangle ABCD can be calculated as:$\\$• Height of the rectangle $= (3P_0 - P_0) = 2P_0 \\$• Width of the rectangle $= (2V_0 - V_0) = V_0 \\$Therefore, the area (work done) is:$\\$Area $=$Height$\times$Width$\\ = 2P_0 \times V_0 \\$However, note the direction of the process:$\\$• The process moves clockwise, indicating positive work done by the system.$\\$• The diagonal paths CB and DA do not contribute to the net work done, as they cancel each other out.$\\$Thus, the net work done by the system in the cycle is:$\\ 0 \\$This corresponds to Option 4.