Answer
$E = (0.208~N/C)~\hat{k}$
Work Step by Step
We can find the electric field at $z = 2.56~cm$ due to a charged disk of $R = 1.80~cm$:
$E = \frac{\sigma}{2\epsilon_0}(1 - \frac{z}{\sqrt{z^2+R^2}})$
$E = \frac{4.50\times 10^{-12}~C/m^2}{(2)(8.854\times 10^{-12}~F/m)}(1 - \frac{0.0256~m}{\sqrt{(0.0256~m)^2+(0.0180~m)^2}})$
$E = 0.04624~N/C$
We can use superposition to find the electric field at $P$:
$E = \frac{\sigma}{2\epsilon} - 0.04624~N/C$
$E = \frac{4.50\times 10^{-12}~C/m^2}{(2)(8.854\times 10^{-12}~F/m)} - 0.04624~N/C$
$E = 0.208~N/C$
Since the charge on the sheet is positive, the electric field at $P$ is directed away from the sheet.
We can express the electric field in unit-vector notation:
$E = (0.208~N/C)~\hat{k}$
