When a proton of charge 'e' mass 'm' is released from rest in a room, it starts to move with initial uniform acceleration $a_0$ toward West. When it is projected towards North with a speed $v_0$, it moves with an initial acceleration 3$a_0$ towards West. The electric and magnetic fields in the room are respectively 

Given data

Charge=e

Mass=m

Acceleration towards west

Two forces =$3m{a}_0$

Projected towards north

Force =F

Concepts used moving charges and magnetism

Explanation

Electric force on the proton 

$F=qE$

Using Newton's second law

$F=ma$

Both forces are equal

$eE=ma$

$E=\frac{m{a}_0}{e}$ along west

For initial given that

$F_1+F_2=3m{a}_0$

$m{a}_0+F_2=3m{a}_0$

$F_2=2m{a}_0$

Maximum magnetic force =$e{v}_{0}B$

$e{v}_{0}B=2m{a}_0$

$B=\frac{2m{a}_0}{e{v}_0}$ down

as force is along west velocity is along north according to 

F=q(v$\times$B)

if west is along -$\hat{i}$ north is along $\hat{j}$ B must be along -$\hat{k}$

so B is downward.