The correct answer is $\boxed{\text{B) 3 mm}}$.
The deflection of a cantilever beam is given by the following formula:
$$\delta = \frac{FL^3}{3EI}$$
where:
- $\delta$ is the deflection,
- $F$ is the applied force,
- $L$ is the length of the beam,
- $E$ is the modulus of elasticity, and
- $I$ is the moment of inertia.
The moment of inertia of a cantilever beam with a uniform cross-section is given by the following formula:
$$I = \frac{bh^3}{12}$$
where:
- $b$ is the width of the beam,
- $h$ is the depth of the beam.
In this case, we are given that $L = 2 \text{ cm}$, $h = 10 \text{ cm}$, $b = 24 \text{ cm}$, and $E = 0.2 \times 10^6 \text{ N/mm}^2$. Substituting these values into the formula for the deflection, we get:
$$\delta = \frac{FL^3}{3EI} = \frac{(100 \text{ N})(2 \text{ cm})^3}{3(0.2 \times 10^6 \text{ N/mm}^2)(\frac{10 \text{ cm}}{12} \text{ mm}^3)} = 3 \text{ mm}$$
Therefore, the deflection of the free end of the cantilever beam is $\boxed{\text{3 mm}}$.
The other options are incorrect because they do not give the correct value for the deflection.