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Given the problem involving two large thin metal plates that are parallel and close to each other with surface charge densities of opposite signs, we are tasked with calculating the electric field intensity \(E\) in two regions:
1. **In the outer region of the first plate**.
2. **Between the plates**.

### Key Information:
- The surface charge density on the inner faces of the plates is given as \( \sigma = 17.7 \times 10^{-10} \, \text{C/m}^2 \).
- The plates are oppositely charged, which means the charges on one plate are positive, and the charges on the other plate are negative.
  
To solve this, we use the concept of **electric field due to an infinite plane of charge**. The electric field due to a single infinite plane with surface charge density \( \sigma \) is given by:

\[
E = \frac{\sigma}{2 \varepsilon_0}
\]

Where:
- \( \varepsilon_0 \) is the permittivity of free space (\( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \)).

### (a) **Electric Field in the Outer Region of the First Plate**:

In the outer region of the first plate, the electric field is only due to the single charged plate, because we assume the second plate is far enough that its influence is negligible in this region. Since the plate has a surface charge density \( \sigma \), the electric field outside the first plate is:

\[
E_{\text{outer}} = \frac{\sigma}{2 \varepsilon_0}
\]

Substituting the values:

\[
E_{\text{outer}} = \frac{17.7 \times 10^{-10}}{2 \times 8.85 \times 10^{-12}} \, \text{N/C}
\]

\[
E_{\text{outer}} = \frac{17.7}{17.7} \times 10^2 = 1 \times 10^2 \, \text{N/C}
\]

Thus, the electric field intensity outside the first plate is:

\[
E_{\text{outer}} = 100 \, \text{N/C}
\]

### (b) **Electric Field Between the Plates**:

In