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Let $\sigma$ be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields in three different region $E_{ I }, E_{ II }$ and $E_{III}$ are

$\vec{E}_{ I }=\frac{2 \sigma}{\epsilon_0} \hat{n}, \vec{E}_{ II }=0, \vec{E}_{ III }=\frac{2 \sigma}{\epsilon_0} \hat{n}$
$\vec{E}_{ I }=0, \vec{E}_{ II }=\frac{\sigma}{\epsilon_0} \hat{n}, \vec{E}_{ III }=0$
$\vec{E}_{ I }=\frac{\sigma}{2 \epsilon_0} \hat{n}, \vec{E}_{\text {II }}=0, \vec{E}_{ III }=\frac{\sigma}{2 \epsilon_0} \hat{n}$
$\vec{E}_{ I }=-\frac{\sigma}{\epsilon_0} \hat{n}, \vec{E}_{\text {II }}=0, \vec{E}_{\text {III }}=\frac{\sigma}{\epsilon_0} \hat{n}$
Solution
Assuming RHS to be $\hat{n}$
$\vec{E}_{ I }=\frac{\sigma}{2 \epsilon_0}(-\hat{n})+\frac{\sigma}{2 \epsilon_0}(-\hat{n})=-\frac{\sigma}{\epsilon_0} \hat{n}$
$\vec{E}_{I I}=0$,
$\vec{E}_{I I I}=\frac{\sigma}{2 \epsilon_0}(\hat{n})+\frac{\sigma}{2 \epsilon_0}(\hat{n})=\frac{\sigma}{\epsilon_0}(\hat{n})$