Step 1: $\sqrt{2}$ is an irrational number (2 is not a perfect square).
Step 2: Use the standard result — the sum of a rational number and an irrational number is always irrational.
Step 3: Assume, if possible, $3+\sqrt{2}=r$ where $r$ is rational. Then $\sqrt{2}=r-3$.
Step 4: The right side $r-3$ is rational, so $\sqrt{2}$ would be rational — a contradiction.
Step 5: Hence $3+\sqrt{2}$ is irrational.
Correct option is (B) an irrational number.