1. Given data

• Specific charge: $\frac{q}{m} = 2 \times 10^{11}\ \text{C/kg}$
• Initial velocity along $x$: $\vec v_0 = 3 \times 10^7\, \hat i\ \text{m/s}$
• Electric field along $y$: $\vec E = 1.8\,\hat j\ \text{kV/m} = 1.8 \times 10^3\,\hat j\ \text{V/m}$
• Length of field region along $x$: $L = 10\ \text{cm} = 0.1\ \text{m}$

We need the deflection in the $y$-direction while the particle is inside the electric field.

2. Acceleration in the $y$-direction

Force due to electric field: $F_y = qE$

So acceleration is $a_y = \frac{F_y}{m} = \frac{qE}{m} = \left(\frac{q}{m}\right)E$

Substitute values: $a_y = (2 \times 10^{11})(1.8 \times 10^3)$ $a_y = 3.6 \times 10^{14}\ \text{m/s}^2$

3. Time spent inside the field

There is no force along $x$, so the $x$-velocity remains constant: $v_x = 3 \times 10^7\ \text{m/s}$

Time to cross the field region: $t = \frac{L}{v_x} = \frac{0.1}{3 \times 10^7}$ $t = \frac{1}{3} \times 10^{-8} = 3.33 \times 10^{-9}\ \text{s}$

4. Deflection in $y$-direction

Initial velocity in $y$ is zero, so $y = \frac{1}{2} a_y t^2$

Now, $t^2 = (3.33 \times 10^{-9})^2 \approx 1.11 \times 10^{-17}$

Hence, $y = \frac{1}{2}(3.6 \times 10^{14})(1.11 \times 10^{-17})$

$y \approx 1.8 \times 1.11 \times 10^{-3}$ $y \approx 2.0 \times 10^{-3}\ \text{m}$

Convert to mm: $2.0 \times 10^{-3}\ \text{m} = 2\ \text{mm}$

5. Final answer

The deflection in the $y$-direction is: $\boxed{2\ \text{mm}}$