The electric field on two sides of a charged sheet is shown in the   figure. The charge density on the sheet is given by

From the figure it is clear that the sheet is placed in an external electric field. Let the electric field due to sheet is E and Eo be the external electric field.

$\begin{array}{l}\therefore E_0+E=12\mathrm{V}/\mathrm{m}\\ E_0-E=8\mathrm{V}/\mathrm{m}\end{array}$

Solving equations,

$E=2\mathrm{V}/\mathrm{m}\text{ and }{E}_0=10\mathrm{V}/\mathrm{m}$

Now electric field due to plate $=\frac{\sigma }{2{\epsilon }_o}=2$

$\therefore \sigma =4{\epsilon }_0$