In finding the electric field using Gauss law the formula $$\left|\vec{E}\right| = \frac{q_{enc}}{\varepsilon_0|A|}$$ is applicable. In the formula $$\varepsilon_0$$ is permittivity of free space, A is the area of the Gaussian surface and $$q_{enc}$$ is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?

We start from the statement of Gauss’s law in integral form

$$\oint_{\text{closed}} \vec{E}\cdot d\vec{A}= \frac{q_{enc}}{\varepsilon_0}.$$

Here $$d\vec{A}$$ is the outward-drawn area element on the chosen Gaussian surface and $$q_{enc}$$ is the total charge enclosed by that surface. The dot product can be written with its magnitude and the angle $$\theta$$ between $$\vec{E}$$ and the outward normal $$\hat{n}$$ to the surface element, giving

$$\oint_{\text{closed}} |\vec{E}|\,\cos\theta \; dA \;=\; \frac{q_{enc}}{\varepsilon_0}.$$

The right side already has no integral, so to convert the left side into the simple product $$|\vec{E}|\,A$$ we must be able to take both $$|\vec{E}|$$ and $$\cos\theta$$ outside the integral sign. This is possible only under the following two simultaneous requirements:

1. $$|\vec{E}|$$ is the same at every point of the Gaussian surface, so that it can be treated as a constant during the integration.

2. The angle $$\theta$$ is the same at every point. The easiest way to secure this is to make $$\theta = 0^{\circ}$$ everywhere, i.e. the electric field is everywhere perpendicular to the surface. When the field is perpendicular everywhere, the scalar potential cannot change along the surface, hence the surface is an equipotential surface.

Under these two conditions we can write

$$\oint_{\text{closed}} |\vec{E}|\,\cos\theta \; dA \;=\; |\vec{E}|\,\cos\theta\, \oint_{\text{closed}} dA \;=\; |\vec{E}|\,(\cos 0^{\circ})\,A \;=\; |\vec{E}|\,A,$$

and Gauss’s law becomes

$$|\vec{E}|\,A = \frac{q_{enc}}{\varepsilon_0}\quad\Longrightarrow\quad |\vec{E}| = \frac{q_{enc}}{\varepsilon_0\,A}.$$

Therefore the compact formula quoted in the question is valid only when (i) the Gaussian surface is an equipotential surface so that $$\vec{E}$$ is normal to it everywhere, and (ii) the magnitude of the field is uniform over the entire surface.

Examining the given options, only Option B states both of these necessary conditions.

Hence, the correct answer is Option B.