Correct Answer - A
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 at point `P` due to two charged plates add up, giving
`E = (sigma)/(2epsilon_(0)) + (sigma)/(2epsilon_(0)) = (sigma)/(epsilon_(0))`
Given, `sigma = 26.4 xx 10^(-12) C//m^(2)`,
`epsilon_(0) = 8.85 xx 10^(-12) C^(2)//N-m^(2)`
Hence, `E = (26.4 xx 10^(-12))/(8.85 xx 10^(-12)) ~~ 3N//C`