Solution:

To calculate the total charge inside the cylinder, we use Gauss's law:

\[
\Phi_E = \frac{q_{\text{inside}}}{\epsilon_0}
\]

Here:
- \(\Phi_E\) is the total electric flux through the surface of the cylinder,
- \(q_{\text{inside}}\) is the total charge enclosed by the cylinder,
- \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/(\text{N·m}^2)\).

Step 1: Electric flux through the cylinder
The cylinder has two flat faces (at \(x = +1\) m and \(x = -1\) m) and a curved surface. The electric field is parallel to the axis, so the flux through the curved surface is **zero**. Thus, only the two flat faces contribute to the flux.

- For the face at \(x = +1\) m:
\(\Phi_E^+ = E \cdot A = 200 \cdot 10^{-2} = 2 \, \text{N·m}^2/\text{C}\),

- For the face at \(x = -1\) m:
\(\Phi_E^- = E \cdot A = -200 \cdot 10^{-2} = -2 \, \text{N·m}^2/\text{C}\).

The total flux is:

\[
\Phi_E = \Phi_E^+ + \Phi_E^- = 2 - (-2) = 4 \, \text{N·m}^2/\text{C}
\]

Step 2: Total charge inside the cylinder
Using Gauss's law:

\[
q_{\text{inside}} = \Phi_E \cdot \epsilon_0 = 4 \cdot (8.85 \times 10^{-12})
\]

Simplify:

\[
q_{\text{inside}} = 35.4 \times 10^{-8} \, \text{C}
\]

Thus, the total charge inside the cylinder is:

\[
{35.4 \times 10^{-8} \, \text{C}}
\]