1. Condition for same path

For the electron to move along the same path even after applying the electric field, the net force perpendicular to its velocity must remain zero.

So, electric force must balance magnetic force:

$eE = evB$

Hence,

$E = vB$

2. Find speed of electron from kinetic energy

Given kinetic energy:

$K = 5\,\text{eV} = 5 \times 1.6 \times 10^{-19}\,\text{J} = 8 \times 10^{-19}\,\text{J}$

Using

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

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

Substitute values:

$v = \sqrt{\frac{2 \times 8 \times 10^{-19}}{9 \times 10^{-31}}}$

$v = \sqrt{\frac{16}{9} \times 10^{12}}$

$v = \frac{4}{3} \times 10^6\,\text{m/s}$

3. Calculate electric field

Magnetic field:

$B = 3\,\mu\text{T} = 3 \times 10^{-6}\,\text{T}$

Therefore,

$E = vB = \left(\frac{4}{3} \times 10^6\right)(3 \times 10^{-6})$

$E = 4\,\text{N/C}$

4. Final answer

$\boxed{4}$

The derived answer matches the stored correct answer.