In a photoelectric effect experiment, the emitter and collector...

The maximum kinetic energy of the emitted photoelectrons corresponds to the minimum wavelength of the incident light, $\lambda = 400 \text{ nm}$.

The energy of the incident photon is given by:
$E = \dfrac{hc}{\lambda} = \dfrac{1242 \text{ eV nm}}{400 \text{ nm}} = 3.105 \text{ eV}$

Using Einstein's photoelectric equation, the maximum kinetic energy K_\max is:
K_\max = E - \phi = 3.105 \text{ eV} - 2.39 \text{ eV} = 0.715 \text{ eV}

For the ammeter to register zero current, the magnetic field must be strong enough to prevent even the most energetic photoelectrons from reaching the collector plate. Assuming the limiting case is for electrons emitted perpendicular to the plates, their trajectory will be a circle. To just miss the opposite plate, the radius of this circular path must be equal to the plate separation $d$.

R = \dfrac{mv_\max}{eB} = d

The maximum velocity v_\max is related to K_\max by K_\max = \dfrac{1}{2}mv_\max^2, which gives v_\max = \sqrt{\dfrac{2K_\max}{m}}.

Substituting v_\max into the radius equation:
B = \dfrac{\sqrt{2mK_\max}}{ed}

Substituting the standard values ($m = 9.1 \times 10^{-31} \text{ kg}$, $e = 1.6 \times 10^{-19} \text{ C}$, $d = 0.1 \text{ m}$):
$B = \dfrac{\sqrt{2 \times 9.1 \times 10^{-31} \times 0.715 \times 1.6 \times 10^{-19}}}{1.6 \times 10^{-19} \times 0.1}$

$B = \dfrac{\sqrt{20.82 \times 10^{-50}}}{1.6 \times 10^{-20}}$

$B = \dfrac{4.56 \times 10^{-25}}{1.6 \times 10^{-20}} = 2.85 \times 10^{-5} \text{ T}$

Answer: $2.85 \times 10^{-5} \text{ T}$