A gas is taken through a cycle A arrow B arrow C arrow A , as shown...

The net work done by the gas in a cyclic process is equal to the area enclosed by the cycle on the P-V diagram.

From the given graph, the cycle $A \rightarrow B \rightarrow C \rightarrow A$ forms a right-angled triangle.

The base of the triangle is $\Delta V = (6 - 2) \times 10^{-3}\ \text{m}^3 = 4 \times 10^{-3}\ \text{m}^3$.

The height of the triangle is $\Delta P = (6 - 3) \times 10^5\ \text{Pa} = 3 \times 10^5\ \text{Pa}$.

Since the cycle is clockwise, the net work done is positive.

$\text{Net work done } (W) = \text{Area of } \triangle ABC$

$W = \dfrac{1}{2} \times \text{base} \times \text{height}$

$W = \dfrac{1}{2} \times (4 \times 10^{-3}\ \text{m}^3) \times (3 \times 10^5\ \text{Pa})$

$W = \dfrac{1}{2} \times 12 \times 10^2\ \text{J}$

$W = 600\ \text{J}$

Answer: $600\ \text{J}$