The total Lorentz force $\vec{F}$ acting on the charge is given by:

$$\vec{F} = q\vec{E} + q(\vec{v} \times \vec{B})$$

Given:

$$q = 2.0\ \mu\text{C} = 2.0 \times 10^{-6}\text{ C}$$

$$\vec{v} = 2.0 \times 10^6 \hat{a}_x\text{ m/s}$$

$$\vec{E} = 10^6 \hat{a}_x\text{ V/m}$$

$$\vec{B} = (0.20 \hat{a}_y + 0.40 \hat{a}_z)\text{ T}$$

First, calculate the electric force:

$$\vec{F}_e = q\vec{E} = (2.0 \times 10^{-6}\text{ C}) \times (10^6 \hat{a}_x\text{ V/m}) = 2.0 \hat{a}_x\text{ N}$$

Next, calculate the magnetic force:

$$ \begin{aligned} \vec{F}_m &= q(\vec{v} \times \vec{B}) \\ &= (2.0 \times 10^{-6}) \times \left[ (2.0 \times 10^6 \hat{a}_x) \times (0.20 \hat{a}_y + 0.40 \hat{a}_z) \right] \\ &= 4.0 \times \left[ \hat{a}_x \times (0.20 \hat{a}_y + 0.40 \hat{a}_z) \right] \\ &= 4.0 \times \left[ 0.20 (\hat{a}_x \times \hat{a}_y) + 0.40 (\hat{a}_x \times \hat{a}_z) \right] \end{aligned} $$

Since $\hat{a}_x \times \hat{a}_y = \hat{a}_z$ and $\hat{a}_x \times \hat{a}_z = -\hat{a}_y$:

$$ \begin{aligned} \vec{F}_m &= 4.0 \times (0.20 \hat{a}_z - 0.40 \hat{a}_y) \\ &= -1.6 \hat{a}_y + 0.8 \hat{a}_z\text{ N} \end{aligned} $$

Therefore, the total force is:

$$\vec{F} = \vec{F}_e + \vec{F}_m = (2.0 \hat{a}_x - 1.6 \hat{a}_y + 0.8 \hat{a}_z)\text{ N}$$

Since this vector is not present in options A, B, or C, the correct choice is None of these (Option D).