To determine the dimensions of $\frac{A^2}{B}$, we start with the given potential energy expression:

$$V = \frac{A x^2}{\sqrt{x} + B}$$

The dimensions of potential energy ($V$) are $[ML^2T^{-2}]$. So, we need to consider the dimensions of the other terms in the expression to match these dimensions. Let's break it down step-by-step.

For the term $\sqrt{x}$, where $x$ represents distance:

$\sqrt{x}$ has dimensions $[L^{1/2}]$

Since both terms in the denominator $\sqrt{x} + B$ have to be of the same dimensions for them to be added together, the dimensions of $B$ must also be $[L^{1/2}]$.

Now the expression becomes:

$$V = \frac{A x^2}{L^{1/2}}$$

Thus, the dimensions must be:

$$[V] = \left[\frac{A x^2}{L^{1/2}}\right]$$

Substitute the dimensions of $V$, $x$, and $L$:

$$[ML^2T^{-2}] = \left[\frac{A L^2}{L^{1/2}}\right]$$

Simplifying the right-hand side:

$$[ML^2T^{-2}] = [A L^{3/2}]$$

Therefore, the dimensions of $A$ must be:

$$[A] = [ML^2T^{-2}] \cdot [L^{-3/2}]$$

$$[A] = [ML^{1/2} T^{-2}]$$

We have dimensions of $A$ and $B$. Now we need to find the dimensions of $\frac{A^2}{B}$:

$$[A^2] = [ML^{1/2} T^{-2}]^2$$

$$[A^2] = [M^2 L T^{-4}]$$

Divide by the dimensions of $B$:

$$\left[\frac{A^2}{B}\right] = \left[\frac{M^2 L T^{-4}}{L^{1/2}}\right]$$

$$\left[\frac{A^2}{B}\right] = [M^2 L^{1/2} T^{-4}]$$

Thus, the dimensions of $\frac{A^2}{B}$ are:

Option C: $$\left[M^2 L^{1/2} T^{-4}\right]$$