One mole of an ideal gas undergoes two different cyclic processes I and II, as shown in the $P-V$ diagrams below. In cycle I, processes $a$, $b$, $c$ and $d$ are isobaric, isothermal, isobaric and isochoric, respectively. In cycle II, processes $a'$, $b'$, $c'$ and $d'$ are isothermal, isochoric, isobaric and isochoric, respectively. The total work done during cycle I is $W_I$ and that during cycle II is $W_{II}$. The ratio $W_I/W_{II}$ is ______.

Step 1: Analyze the processes in Cycle I.

Calculate the work done in each individual process of Cycle I.

$$W_a = 4P_0(2V_0 - V_0) = 4P_0V_0$$ $$W_b = \int_{2V_0}^{4V_0} \frac{8P_0V_0}{V} \, dV = 8P_0V_0 \ln 2$$ $$W_c = 2P_0(V_0 - 4V_0) = -6P_0V_0$$ $$W_d = 0$$

Step 2: Calculate the total work done in Cycle I ($W_I$).

Sum the work done in all four processes of Cycle I.

$$W_I = W_a + W_b + W_c + W_d$$ $$W_I = 4P_0V_0 + 8P_0V_0 \ln 2 - 6P_0V_0$$ $$W_I = (8\ln 2 - 2)P_0V_0$$

Step 3: Analyze the processes in Cycle II.

Calculate the work done in each individual process of Cycle II.

$$W_{a'} = \int_{V_0}^{2V_0} \frac{4P_0V_0}{V} \, dV = 4P_0V_0 \ln 2$$ $$W_{b'} = 0$$ $$W_{c'} = P_0(V_0 - 2V_0) = -P_0V_0$$ $$W_{d'} = 0$$

Step 4: Calculate the total work done in Cycle II ($W_{II}$).

Sum the work done in all four processes of Cycle II.

$$W_{II} = W_{a'} + W_{b'} + W_{c'} + W_{d'}$$ $$W_{II} = (4\ln 2 - 1)P_0V_0$$

Step 5: Find the ratio of the work done in the two cycles.

Divide the total work of Cycle I by the total work of Cycle II.

$$\frac{W_I}{W_{II}} = \frac{(8\ln 2 - 2)P_0V_0}{(4\ln 2 - 1)P_0V_0}$$ $$\frac{W_I}{W_{II}} = \frac{2(4\ln 2 - 1)}{4\ln 2 - 1}$$ $$\frac{W_I}{W_{II}} = 2$$

Answer: 2