Step 1: Remember that water can react too.
During electrolysis of a salt solution there is a competition at each electrode between the salt ions and water itself. Whichever species is easier to discharge is the one that reacts. The phrase 'very dilute' is the big clue here.
Step 2: Look at the cathode.
Sodium ions are extremely reluctant to gain electrons, so water is reduced in their place and hydrogen gas comes off:
\[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \]
Step 3: Look at the anode.
Chloride could in principle be oxidised, but a very dilute solution barely has any chloride present. So water is oxidised instead and oxygen gas is released:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
Step 4: Read off the result.
We end up with hydrogen at the cathode and oxygen at the anode, so the statement about oxygen at the anode is the right one.
\[ \boxed{\text{Option (C): } O_2 \text{ is evolved at the anode}} \]