A positive charge particle of 100   mg is thrown in opposite direction to a uniform electric field of…

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

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

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

Using, $v^2=u^2-2as$

We have, $0^2=u^2-2as$

$s=\frac{v^2}{2a}=\frac{{\left(200\right)}^2}{2\times 4\times {10}^4}=0.5 \mathrm{m}$