The electric field on two sides of a charged plate is shown in the figure. If the charge density on the plate is $n{\epsilon }_0$, then n is

The electric field produced by thin charged sheet is given by  $\frac{\sigma }{2{\epsilon }_0}$

[ $\sigma$ = surface charge density]
$\therefore$ An external electric field is also present so we will use principal of superposition to find net electric field.
From the figure, it is clear that the plate is placed in an external electric field. Let the electric field due to plate is E and $E_0$ be the external electric field.
$E_0+E=12V/m$   …(i)
$E_0-E=8V/m$   …(ii)
Solving Eqs. (i) and (ii), we get E = 2 V/m
Now, electric field due to plate  $=\frac{\sigma }{2{\in }_0}=2$   then $\sigma =4{\in }_0\Rightarrow$ n = 4