1. Let the vector have magnitude $V$ and make an angle of $30^\circ$ with the $y$-axis.

2. Since the angle is given with the $y$-axis: $|V_y| = V\cos 30^\circ$ and $|V_x| = V\sin 30^\circ$

3. Given: $|V_y| = 2\sqrt{3}$ So, $V\cos 30^\circ = 2\sqrt{3}$

4. Use $\cos 30^\circ = \frac{\sqrt{3}}{2}$: $V\left(\frac{\sqrt{3}}{2}\right) = 2\sqrt{3}$ $V = \frac{2\sqrt{3} \cdot 2}{\sqrt{3}} = 4$

5. Now find the $x$-component: $|V_x| = V\sin 30^\circ = 4\left(\frac{1}{2}\right) = 2$

6. Hence, the magnitude of the $x$-component is: $\boxed{2}$

7. Checking options:

   • A: $\sqrt{3}$
   • B: $2$ ✅
   • C: $6$
   • D: $\frac{1}{\sqrt{3}}$

Therefore, the correct option is B.