An ideal gas undergoes a clockwise cyclic process consisting of...

The cyclic process forms a right-angled triangle on the $P-V$ diagram.

The isobaric expansion occurs at a constant pressure of $2P_0$ from volume $V_0$ to $3V_0$.
Work done during this expansion, W_\exp = P \Delta V = 2P_0(3V_0 - V_0) = 4P_0V_0.

The net work done in the clockwise cycle is the area of the triangle enclosed on the $P-V$ diagram.
Base of the triangle (change in volume) = $3V_0 - V_0 = 2V_0$.
Height of the triangle (change in pressure) = $2P_0 - P_0 = P_0$.
Net work, $W_{net} = \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 2V_0 \times P_0 = P_0V_0$.

The required ratio is \dfrac{W_\exp}{W_{net}} = \dfrac{4P_0V_0}{P_0V_0} = 4.

Answer: $4$