1. Electric field due to an infinite plane sheet

For a uniformly charged infinite plane sheet with surface charge density $+\sigma$,

$E = \frac{\sigma}{2\varepsilon_0}$

The field is directed away from the positively charged sheet.

2. Force and acceleration of the electron

An electron has charge $-e$, so the force on it is toward the sheet.

Magnitude of force:

$F = eE = e\cdot \frac{\sigma}{2\varepsilon_0}$

Hence acceleration magnitude is

$a = \frac{F}{m} = \frac{e\sigma}{2m\varepsilon_0}$

This acceleration is toward the sheet.

3. Choose coordinates

Let the sheet be at $x=0$ and the electron initially be at $x=1\,\text{m}$.

To get the maximum possible $\sigma$ while still striking the sheet at $t=1\,\text{s}$, the electron must initially move away from the sheet with speed $1\,\text{m/s}$. If it initially moved toward the sheet, a larger acceleration would make it arrive earlier, so the allowed $\sigma$ would be smaller.

Thus,

$x_0 = 1, \quad u = +1\,\text{m/s}$

Acceleration is toward the sheet, so

$a = -\frac{e\sigma}{2m\varepsilon_0}$

The electron strikes the sheet at $t=1\,\text{s}$, so $x=0$ at $t=1$.

Using

$x = x_0 + ut + \frac{1}{2}at^2$

we get

$0 = 1 + (1)(1) + \frac{1}{2}a(1)^2$

$0 = 2 + \frac{a}{2}$

$a = -4\,\text{m/s}^2$

So the required acceleration magnitude is

$|a| = 4$

4. Find $\sigma$

From

$\frac{e\sigma}{2m\varepsilon_0} = 4$

$\sigma = \frac{8m\varepsilon_0}{e}$

Comparing with

$\sigma = \alpha\left[\frac{m\varepsilon_0}{e}\right] \frac{C}{m^2}$

we get

$\alpha = 8$

5. Final answer

$\boxed{8}$