Separated by a distance of 2.00 cm , two large conducting plates...

Given the distance between the plates, $d = 2.00 \text{ cm} = 0.02 \text{ m}$, and the time taken by the electron, $t = 2.00 \text{ \mu s} = 2.00 \times 10^{-6} \text{ s}$.

Using the kinematic equation for a particle starting from rest:
$d = \dfrac{1}{2} a t^2$

Solving for acceleration $a$:
$a = \dfrac{2d}{t^2} = \dfrac{2 \times 0.02}{(2.00 \times 10^{-6})^2} = \dfrac{0.04}{4.00 \times 10^{-12}} = 10^{10} \text{ m/s}^2$

The force experienced by the electron in the uniform electric field $E$ is $F = eE = m_e a$.
Thus, the electric field between the plates is:
$E = \dfrac{m_e a}{e} = \dfrac{9.11 \times 10^{-31} \times 10^{10}}{1.60 \times 10^{-19}} = 5.69 \times 10^{-2} \text{ V/m}$

For two large parallel conducting plates with opposite charges, the electric field in the region between them is given by:
$E = \dfrac{\sigma}{\epsilon_0}$
where $\sigma$ is the surface charge density on the inner surfaces.

Solving for $\sigma$:
$\sigma = \epsilon_0 E = (8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2) \times (5.69 \times 10^{-2} \text{ V/m})$
$\sigma \approx 5.04 \times 10^{-13} \text{ C/m}^2$

The closest matching option is $5.05 \times 10^{-13} \text{ C/m}^2$.

Answer: $5.05 \times 10^{-13} \text{ C/m}^2$