When a proton is released from rest in a room, it starts with an initial acceleration a₀ towards west. When it is projected towards north with a speed v₀ it moves with an initial acceleration 3a₀ towards west. The electric and magnetic fields in the room are respectively.

when the velocity is zero, it will have only electric force.

$\overrightarrow{{\mathrm{F}}_{\mathrm{e}}}=\mathrm{e}\overrightarrow{\mathrm{E}}=\mathrm{m}\overrightarrow{{\mathrm{a}}_0}$

$\overrightarrow{\mathrm{E}}=\frac{\mathrm{m}\overrightarrow{{\mathrm{a}}_0}}{\mathrm{e}}$

Acceleration is towards west, so electric field is towards west.

When velocity is not zero, the magnetic force may act on the electron. 

Here, increases acceleration is due to magnetic force.

$\overrightarrow{{\mathrm{F}}_{\mathrm{m}}}=\mathrm{m}(3\overrightarrow{{\mathrm{a}}_0}-\overrightarrow{{\mathrm{a}}_0})=2\mathrm{m}\overrightarrow{{\mathrm{a}}_0}=\mathrm{e}(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}})$

As velocity is in north direction and acceleration is in west direction, the magnetic field will be in vertically downward direction.

$\Rightarrow 2{\mathrm{ma}}_0={\mathrm{ev}}_0\mathrm{B}$

$\Rightarrow \mathrm{B}=\frac{2{\mathrm{ma}}_0}{{\mathrm{ev}}_0}$