The correct answer is $\boxed{\text{A}. 0.56 \times 10^{15} \text{ m/s}^2}$.
The acceleration of an electron in a uniform electric field is given by:
$$a = \frac{E}{m}$$
where $E$ is the electric field strength and $m$ is the mass of the electron.
The electric field strength between two parallel plates is given by:
$$E = \frac{\Delta V}{d}$$
where $\Delta V$ is the potential difference between the plates and $d$ is the distance between the plates.
In this case, we are given that $\Delta V = 100 \text{ V}$ and $d = 10 \text{ mm} = 0.01 \text{ m}$. Substituting these values into the equation for the electric field strength, we get:
$$E = \frac{100 \text{ V}}{0.01 \text{ m}} = 10^4 \text{ V/m}$$
The mass of an electron is $9.11 \times 10^{-31} \text{ kg}$. Substituting this value into the equation for the acceleration, we get:
$$a = \frac{10^4 \text{ V/m}}{9.11 \times 10^{-31} \text{ kg}} = 0.56 \times 10^{15} \text{ m/s}^2$$
Therefore, the acceleration of an electron placed between two infinite parallel plates 10 mm apart with a potential difference of 100 V is $0.56 \times 10^{15} \text{ m/s}^2$.
Option A is the correct answer because it is the only option that is within the range of values that are expected for the acceleration of an electron in a uniform electric field. Options B, C, and D are incorrect because they are too large.