The problem asks to simplify the expression $2\sqrt{3} - \frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}}$.

1. Problem Description

The problem asks to simplify the expression $2\sqrt{3} - \frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}}$.

2. Solution Steps

First, we simplify the second term by rationalizing the denominator:

$$\frac{6}{\sqrt{3}} = \frac{6}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$$

Next, we simplify the third term. We know that $27 = 9 \cdot 3$, so $\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}$. Therefore,

$$\frac{3}{\sqrt{27}} = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3} = \frac{1}{3}\sqrt{3}$$

Now, we substitute these simplified terms back into the original expression:

$2\sqrt{3} - \frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}} = 2\sqrt{3} - 2\sqrt{3} + \frac{1}{3}\sqrt{3}$

Combining like terms, we get:

$2\sqrt{3} - 2\sqrt{3} + \frac{1}{3}\sqrt{3} = 0\sqrt{3} + \frac{1}{3}\sqrt{3} = \frac{1}{3}\sqrt{3}$

3. Final Answer

The simplified expression is $\frac{1}{3}\sqrt{3}$.

The answer is B.