To determine the dimensions of $$\epsilon_{\mathrm{o}} \mathrm{E}^2$$, we need to first understand the dimensional formulas of each component in the expression.

1. Permittivity of free space, $$\epsilon_{\mathrm{o}}$$:

The permittivity of free space has the dimensions: $$\left[\epsilon_{\mathrm{o}}\right] = [\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2]$$.

2. Electric field, $$\mathrm{E}$$:

The electric field $$\mathrm{E}$$ has the dimensions: $$\left[\mathrm{E}\right] = [\mathrm{M} \mathrm{L} \mathrm{T}^{-3} \mathrm{A}^{-1}]$$.

Now, let's calculate the dimensions of $$\epsilon_{\mathrm{o}} \mathrm{E}^2$$:

$\begin{equation} \left[\epsilon_{\mathrm{o}} \mathrm{E}^2 \right] = \left[\epsilon_{\mathrm{o}}\right] \left[\mathrm{E}\right]^2 = \left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2\right] \left([\mathrm{M} \mathrm{L} \mathrm{T}^{-3} \mathrm{A}^{-1}]^2\right] = \left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2\right] \left([\mathrm{M}^2 \mathrm{L}^2 \mathrm{T}^{-6} \mathrm{A}^{-2}]\right] \end{equation}$

Combining the dimensions:

$\begin{equation} \left[\epsilon_{\mathrm{o}} \mathrm{E}^2 \right] = \left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2\right] \left[\mathrm{M}^2 \mathrm{L}^2 \mathrm{T}^{-6} \mathrm{A}^{-2}\right] = \left[\mathrm{M}^{-1 + 2} \mathrm{L}^{-3 + 2} \mathrm{T}^{4 - 6} \mathrm{A}^{2 - 2}\right] = \left[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}\right] \end{equation}$

Therefore, the dimensions of $$\epsilon_{\mathrm{o}} \mathrm{E}^2$$ are $$\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]$$, which corresponds to Option A.

Hence, the correct answer is Option A: $$\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]$$.