Entropy change for an irreversible isolated system is

Answer: Option D

From second law of thermodynamics
$$\begin{array}{rl}& TdS⩾\delta Q\\ & \Rightarrow TdS⩾dU+\delta W\end{array}$$
For an irreversible process $$S-dU-\delta W>0.$$    So, system undergoing irreversible change under constant internal energy and constant volume will have entropy change greater than zero $$dS>0.$$
And for an reversible process $$TdS-dU-\delta W=0.$$