The correct option is B 13R/6
W=vf∫viPdV
Work done is area under PV graph, ie area of trapezium,hence
W=12(6P0+3P0)4V0
W=18P0V0
ΔU=PfVf−PiVi5/3−1=30P0V0−3P0V02/3
=81P0V02
ΔQ=W+ΔU=1172P0V0
nC(Tf−Ti)=1172P0V0
C(6P0×5V0R−3P0V0R)=1172P0V0
C=1172P0V027P0V0R=13R6