Yes, direct formulas can simplify the solution for problems like this involving conducting plates. Here's how:

Formula for charge redistribution in conductors:
For three parallel conducting plates with charges \( Q_1, Q_2, Q_3 \):
1. The charge on the inner faces is:
\[
Q_{\text{inner}} = \frac{Q_1 - Q_3}{2}
\]
2. The charge on the  outer faces remains equal to the net charge on the respective plates:
- Outer left face: \( Q_1 \)
- Outer right face: \( Q_3 \)

---

Applying the formula:
- Given charges:
\[
Q_1 = -2Q, \, Q_2 = Q, \, Q_3 = -Q
\]
- Charge on the inner faces of plate \( b \):
\[
Q_{\text{inner}} = \frac{Q_1 - Q_3}{2} = \frac{-2Q - (-Q)}{2} = \frac{-2Q + Q}{2} = \frac{-Q}{2}
\]

- Since the **right face of \( b \)** contributes to the inner faces:
\[
Q_{\text{right face of \( b \)}} = 0
\]

Final Answer:
\[
{Zero (0)}
\]

The question states the mathematical form of Gauss's law as:

\[
\varepsilon_{0}\oint \overrightarrow{E} \cdot d\overrightarrow{S} = q
\]

Where:
- \(\varepsilon_0\) is the permittivity of free space,
- \(\overrightarrow{E}\) is the electric field vector,
- \(d\overrightarrow{S}\) is the infinitesimal area vector on the Gaussian surface,
- \(q\) is the total charge enclosed within the Gaussian surface.

Key Points in Gauss's Law:

1. The electric flux through the Gaussian surface (\(\oint \overrightarrow{E} \cdot d\overrightarrow{S}\)) depends only on the charge enclosed (\(q\)) within the surface.

2. The electric field \(\overrightarrow{E}\) at any point on the Gaussian surface depends on all charges in the system—both inside and outside the Gaussian surface.

Why Does \(E\) Depend on Charges Outside the Gaussian Surface?

While Gauss's law calculates flux based only on enclosed charge, the electric field \(\overrightarrow{E}\) at a point on the Gaussian surface is influenced by all charges, regardless of their location (inside or outside the surface). Here's why:

- Charges inside the Gaussian surface: These contribute directly to the net flux as per Gauss's law.
- Charges outside the Gaussian surface: These do not contribute to the net flux (their contributions cancel out overall due to symmetry), but they **still influence the local value of \(\overrightarrow{E}\)**.

Example:

- Imagine a spherical Gaussian surface around a point charge \(q_1\). If another charge \(q_2\) is placed outside the sphere, it doesn't affect the total flux, but it does contribute to the electric field at various points on the sphere.

Thus, the electric field \(\overrightarrow{E}\) depends on all charges in the vicinity, while the total flux (as per Gauss's law) depends solely on the charges enclosed. This is why the correct answer is:

"E depends on the charge which is inside and outside the Gaussian surface."