Emitted from an electron gun with almost zero velocity, electrons...

The wavelength of the emitted red light is $656.3 \text{ nm}$, which corresponds to the Balmer series transition from $n = 3$ to $n = 2$.

For the hydrogen atom to emit this light, it must first be excited from its ground state ($n = 1$) to at least the $n = 3$ state.

The energy required for this excitation is:

$\Delta E = 13.6 \left( \dfrac{1}{1^2} - \dfrac{1}{3^2} \right) \text{ eV}$

$\Delta E = 13.6 \times \dfrac{8}{9} \text{ eV} \approx 12.1 \text{ eV}$

The kinetic energy gained by an electron accelerated by an electric field $E$ over a distance $d$ is given by $K = eEd$.

To excite the hydrogen atom, the kinetic energy of the electron must be at least equal to the excitation energy:

$eEd = 12.1 \text{ eV}$

Given $d = 1.0 \text{ m}$, we have:

$e \times E \times 1.0 = 12.1 \text{ eV}$

$E = 12.1 \text{ V/m}$