let angle of deviation is a.

from figure, sina $=\mathrm{d} / \mathrm{R}$

where, $\mathrm{d}=3 \mathrm{mv} / 5 \mathrm{qb}$ and we know radius of circular path attained by charged particle $, \mathrm{R}=\mathrm{mv} / \mathrm{qb}$

now, sina $=(3 m v / 5 q b) /(m v / q b)=3 / 5=\sin 37^{\circ}$

$\Rightarrow a=37^{\circ}$

we know, angle of deviation = emergent angle - incidence angle

$=\theta+53^{\circ}$

or, $37^{\circ}=\theta+53^{\circ}$

or, $\theta=90^{\circ}$