A positive charge particle of 100mg is thrown in opposite direction...

Force experienced by a particle with charge $q$ in an electric field $F = qE$.

Now, the acceleration produced is given as $a = \dfrac{F}{m} = \dfrac{qE}{m}$$= \dfrac{40 \times {10}^{- 6} \times {10}^{5}}{1 \times {10}^{- 4}}$

$= 4 \times {10}^{4}m{s}^{- 2}$ (As the particle is projected against the electric field, hence it is decelerated)

Using, ${v}^{2} = {u}^{2} - 2as$

We have, ${0}^{2} = {u}^{2} - 2as$

$s = \dfrac{{v}^{2}}{2a} = \dfrac{{\left(200\right)}^{2}}{2 \times 4 \times {10}^{4}} = 0 . 5m$