Step 1: Use the cyclic process rule. Over one full cycle the gas returns to its starting state, so the internal energy change is $\Delta U = 0$. Step 2: Relate heat to work. By the first law, $Q_{net} = \Delta U + W_{net} = W_{net}$. So the heat supplied equals the net work done. Step 3: Work equals enclosed area. On a $P$-$V$ diagram, the net work in a cycle equals the area enclosed by the loop. Step 4: Read the rectangle's sides. From the graph, the pressure spans $\Delta P = 300 - 100 = 200$ N/m$^2$ and the volume spans $\Delta V = 5 - 2 = 3$ m$^3$. Step 5: Compute the area. $W = \Delta P \times \Delta V = 200 \times 3 = 600$ J. Step 6: Get the heat supplied. Since $Q = W$, the total heat supplied is $600$ J, which is option D. \[ \boxed{ 600 \text{ J} } \]
