1. We need to evaluate

$$L=\lim_{x\to 3}\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt2}.$$

2. Direct substitution gives

$$\sqrt{3(3)}-3=\sqrt9-3=0,$$

and

$$\sqrt{2(3)-4}-\sqrt2=\sqrt2-\sqrt2=0.$$

So the limit is of the indeterminate form $\frac{0}{0}$.

3. Rationalize both numerator and denominator:

For the numerator,

$$\sqrt{3x}-3=\frac{(\sqrt{3x}-3)(\sqrt{3x}+3)}{\sqrt{3x}+3} =\frac{3x-9}{\sqrt{3x}+3} =\frac{3(x-3)}{\sqrt{3x}+3}.$$

For the denominator,

$$\sqrt{2x-4}-\sqrt2 =\frac{(\sqrt{2x-4}-\sqrt2)(\sqrt{2x-4}+\sqrt2)}{\sqrt{2x-4}+\sqrt2} =\frac{(2x-4)-2}{\sqrt{2x-4}+\sqrt2} =\frac{2(x-3)}{\sqrt{2x-4}+\sqrt2}.$$

4. Substitute these into the limit:

$$L=\lim_{x\to 3} \frac{\frac{3(x-3)}{\sqrt{3x}+3}}{\frac{2(x-3)}{\sqrt{2x-4}+\sqrt2}}.$$

5. Cancel the common factor $(x-3)$:

$$L=\lim_{x\to 3} \frac{3}{\sqrt{3x}+3}\cdot \frac{\sqrt{2x-4}+\sqrt2}{2}.$$

6. Now substitute $x=3$:

$$L=\frac{3}{\sqrt9+3}\cdot \frac{\sqrt2+\sqrt2}{2} =\frac{3}{3+3}\cdot \frac{2\sqrt2}{2} =\frac{3}{6}\cdot \sqrt2 =\frac{\sqrt2}{2}.$$

7. Simplify:

$$\frac{\sqrt2}{2}=\frac{1}{\sqrt2}.$$

Hence,

$$\boxed{\frac{1}{\sqrt2}}.$$

So the correct option is B.