Step 1: Split the fraction term by term.
$$\frac{\sqrt{3}-3}{\sqrt{3}}=\frac{\sqrt{3}}{\sqrt{3}}-\frac{3}{\sqrt{3}}$$
Step 2: Simplify each part. $\dfrac{\sqrt{3}}{\sqrt{3}}=1$ and $\dfrac{3}{\sqrt{3}}=\dfrac{3}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}=\sqrt{3}$.
Step 3: Therefore
$$\frac{\sqrt{3}-3}{\sqrt{3}}=1-\sqrt{3}$$
Step 4: Since 1 is rational and $\sqrt{3}$ is irrational, their difference $1-\sqrt{3}$ is irrational.
Correct option is (B) an irrational number.