In a vacuum chamber, a particle of charge 1\ and mass 1 mg is...

For the time interval $0 \le t \le 0.2$ s, the particle is under the influence of the electric field and gravity.
The charge-to-mass ratio of the particle is $\dfrac{q}{m} = \dfrac{10^{-6}}{10^{-6}} = 1$ C kg$^{-1}$.

The acceleration of the particle is:
$\vec{a} = \dfrac{q\vec{E}}{m} + \vec{g} = 1(\hat{i}) - 10\hat{j} = \hat{i} - 10\hat{j}$ ms$^{-2}$

Given the initial velocity $\vec{v}_0 = \hat{i} + 2\hat{j}$ ms$^{-1}$ and initial vertical position $y_0 = 0$ (since it is projected from the $XZ$ plane), we can find the velocity and position at $t = 0.2$ s.

Velocity at $t = 0.2$ s:
$\vec{v}(0.2) = \vec{v}_0 + \vec{a}t = (\hat{i} + 2\hat{j}) + (\hat{i} - 10\hat{j})(0.2) = 1.2\hat{i}$ ms$^{-1}$

Vertical position at $t = 0.2$ s:
$y(0.2) = y_0 + v_{0y}t + \dfrac{1}{2}a_y t^2 = 0 + 2(0.2) - \dfrac{1}{2}(10)(0.2)^2 = 0.4 - 0.2 = 0.2$ m $= 20$ cm

For $t > 0.2$ s, the electric field is switched off and the magnetic field $\vec{B} = 6\hat{j}$ T is switched on. The magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ acts only in the $XZ$ plane because $\vec{B}$ is along the $y$-axis. The vertical motion is only affected by gravity.

Let $t' = t - 0.2$ be the time elapsed after the fields are switched. The initial vertical velocity for this phase is $v_y(0.2) = 0$.
The vertical position as a function of $t'$ is:
$y(t') = y(0.2) + v_y(0.2)t' - \dfrac{1}{2}gt'^2 = 0.2 - 5t'^2$

Evaluating the options:
At $t = 0.3$ s ($t' = 0.1$ s):
$y(0.1) = 0.2 - 5(0.1)^2 = 0.2 - 0.05 = 0.15$ m $= 15$ cm. (Option A is correct)

At $t = 0.4$ s ($t' = 0.2$ s):
$y(0.2) = 0.2 - 5(0.2)^2 = 0.2 - 0.2 = 0$ m $= 0$ cm. (Option B is incorrect)

At $t = 0.35$ s ($t' = 0.15$ s):
$y(0.15) = 0.2 - 5(0.15)^2 = 0.2 - 0.1125 = 0.0875$ m $\neq 0$. (Option D is incorrect)

For the radius of the trajectory for $t > 0.2$ s, the particle moves in a helical path. The radius of the circular projection in the $XZ$ plane is determined by the velocity perpendicular to the magnetic field, $v_{\perp} = 1.2$ ms$^{-1}$.
$R = \dfrac{mv_{\perp}}{qB} = \dfrac{10^{-6} \times 1.2}{10^{-6} \times 6} = 0.2$ m $= 20$ cm. (Option C is correct)

Answer: The vertical distance of the particle from the $XZ$ plane at $t = 0.3$ s is $15$ cm.; The radius of the trajectory of the particle for $t > 0.2$ s is $20$ cm.