An ideal monatomic gas (n moles) undergoes the cycle A→ B→ C→ A on a P-V diagram: A(P0,V0)→ B(2P0,V0) at constant volume, B→ C(2P0,2V0) at constant…

Net work is the enclosed area of the right triangle with vertical side $P_0$ and horizontal side $V_0$: $W=\tfrac12P_0V_0$, so (A) is correct. For a monatomic gas $U=\tfrac32PV$. On $A\to B$: $Q=\Delta U=\tfrac32P_0V_0$. On $B\to C$: $Q=\tfrac52\Delta(PV)=\tfrac52(4-2)P_0V_0=5P_0V_0$. Heat absorbed per cycle is therefore $6.5P_0V_0$, not $6P_0V_0$, so (B) is wrong. The line $C\to A$ has $P=P_0V/V_0$, giving $dQ=dU+PdV=4P_0V\,dV/V_0$, which is negative for every $dV<0$: heat is rejected throughout, so (C) is wrong (total $Q_{CA}=-6P_0V_0$, consistent with $W_{net}=6.5-6$). Along this line $nR\,dT=2P_0V\,dV/V_0$, hence $C=dQ/(n\,dT)=2R$, not $3R$, so (D) is wrong.