Equate potential energy to kinetic energy.

The potential energy (PE) lost by an electron with charge $e$ moving through a potential difference $V$ is $PE = eV$.
The kinetic energy (KE) gained by an electron with mass $m$ and velocity $v$ is $KE = \frac{1}{2}mv^2$.
According to the conservation of energy:

$$eV = \frac{1}{2}mv^2$$

Multiply both sides by 2:

$$2eV = mv^2$$

Divide by $m$:

$$v^2 = \frac{2eV}{m}$$

Take the square root of both sides:

$$v = \sqrt{\frac{2eV}{m}}$$

This can be rewritten using the given charge-to-mass ratio $\frac{e}{m}$:

$$v = \sqrt{2V \left(\frac{e}{m}\right)}$$

Given:
Potential difference, $V = 3000 \text{ V}$
Ratio of charge to mass, $\frac{e}{m} = 1.8 \times 10^{11} \text{ Ckg}^{-1}$

$$v = \sqrt{2 \times 3000 \text{ V} \times 1.8 \times 10^{11} \text{ Ckg}^{-1}}$$ $$v = \sqrt{6000 \times 1.8 \times 10^{11}}$$ $$v = \sqrt{10800 \times 10^{11}}$$ $$v = \sqrt{10.8 \times 10^3 \times 10^{11}}$$ $$v = \sqrt{10.8 \times 10^{14}}$$

$$v = \sqrt{10.8} \times \sqrt{10^{14}}$$ $$v \approx 3.286 \times 10^7 \text{ ms}^{-1}$$

Rounding to two significant figures, we get:

$$v \approx 3.3 \times 10^7 \text{ ms}^{-1}$$

A. $3.3 \times 10^7 \text{ms}^{-1}$
B. $1.0 \times 10^7 \text{ms}^{-1}$
C. $3.3 \times 10^4 \text{ms}^{-1}$
D. $2.2 \times 10^7 \text{ms}^{-1}$

The calculated velocity matches option A.

The final answer is $\boxed{\text{A. } 3.3 \times 10^7 \text{ms}^{-1}}$.