Explanation

(c) The situation is shown in the figure. Plate 1 has surface charge density $\sigma$ and plate 2 has surface charge density$–\sigma$. The electric fields
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at point $P$ due to two charged plates add up, giving
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \frac{\sigma}{2\varepsilon_{0}} + \frac{\sigma}{2\varepsilon_{0}} = \frac{\sigma}{\varepsilon_{0}}$
Given , $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \sigma = 26.4 \times 10^{- 12}{Cm}^{- 2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \varepsilon_{0} = 8.85 \times 10^{- 12}C^{2}N^{- 1}m^{- 2}$
Hence , $E\ = \ \frac{26.4 \times 10^{- 12}}{8.85 \times 10^{- 12}}\ \approx 3{NC}^{- 1}$
Note the direction of electric field is from the positive to the negative plate.