One mole of an ideal gas undergoes two different cyclic processes I...

For first case:

${W}_{I} = {W}_{a} + {W}_{b} + {W}_{c} + {W}_{d}$

$= 4{P}_{0}\left(2{V}_{0} - {V}_{0}\right) + nRTln\left(\dfrac{4{V}_{0}}{2{V}_{0}}\right) + 2{P}_{0}\left({V}_{0} - 4{V}_{0}\right) + 0$

$= 4{P}_{0}{V}_{0} + nR\left(\dfrac{8{P}_{0}{V}_{0}}{nR}\right)\ln 2 - 6{P}_{0}{V}_{0}$

$= 8{P}_{0}{V}_{0}\ln 2 - 2{P}_{0}{V}_{0}$

For second case:

${W}_{II} = W{'}_{a} + W{'}_{b} + W{'}_{c} + W{'}_{d}$

$= nRTln\left(\dfrac{2{V}_{0}}{{V}_{0}}\right) + 0 + {P}_{0}\left({V}_{0} - 2{V}_{0}\right) + 0$

$= nR\left(\dfrac{4{P}_{0}{V}_{0}}{nR}\right)\ln 2 - {P}_{0}{V}_{0}$

$= 4{P}_{0}{V}_{0}\ln 2 - {P}_{0}{V}_{0}$

Therefore, required ratio is $\dfrac{{W}_{I}}{{W}_{II}} = \dfrac{8{P}_{0}{V}_{0}\ln 2 - 2{P}_{0}{V}_{0}}{4{P}_{0}{V}_{0}\ln 2 - {P}_{0}{V}_{0}} = 2$