The line connecting A and B passes through the origin, which is an isobaric process, TB = 2ta, external work Q 1=(VB-VA)*P, and thermal increment Q2 = (TB-TA) × CP.
The isovolumetric processes of B and C do no external work, and give off heat Q3 = (Tb-Ta) × CV.
C, the external work P = RTA/V of a process isothermal process on gas is integrated to obtain RTAlN (VC/VA) = RTAlN2.
The external work on gas is RTA LN2-(VB-VA) * p = RTA (LN2-1) < 0.
This is a positive cycle.
The heat engine efficiency formula η t = w/q1= (q1-Q2)/q1(irrelevant to this question).
The total heat absorbed is qab = ta× CP = 2.5 RTA.
The total heat released is 1.5RTA.
Efficiency = (2.5- 1.5)/2.5 = 40%
Give it to me, huh?