The electric field in a certain region is defined by E = (E_0...

The electric field is given by $\vec{E} = \dfrac{E_0 x}{l} \hat{i}$.

The flux through the cubical volume is non-zero only for the faces perpendicular to the x-axis, as the electric field has only an x-component.

At the face $x = 0$, the electric field is zero, so the flux $\Phi_1 = 0$.

At the face $x = a$, the electric field is $\vec{E} = \dfrac{E_0 a}{l} \hat{i}$. The area vector for this face is $\vec{A} = a^2 \hat{i}$.

The flux through this face is $\Phi_2 = \vec{E} \cdot \vec{A} = \dfrac{E_0 a^3}{l}$.

The total electric flux through the cube is $\Phi = \Phi_1 + \Phi_2 = \dfrac{E_0 a^3}{l}$.

According to Gauss's law, the total flux is related to the enclosed charge by $\Phi = \dfrac{q_{\text{enclosed}}}{\varepsilon_0}$.

Therefore, $q_{\text{enclosed}} = \varepsilon_0 \dfrac{E_0 a^3}{l}$.

Substituting the given values $E_0 = 5 \times 10^3 \text{ N C}^{-1}$, $l = 2 \times 10^{-2} \text{ m}$, $a = 1 \times 10^{-2} \text{ m}$, and $\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}$:

$q_{\text{enclosed}} = 8.85 \times 10^{-12} \times \dfrac{5 \times 10^3 \times (10^{-2})^3}{2 \times 10^{-2}}$

$q_{\text{enclosed}} = 8.85 \times 10^{-12} \times \dfrac{5 \times 10^{-3}}{2 \times 10^{-2}}$

$q_{\text{enclosed}} = 8.85 \times 10^{-12} \times 0.25$

$q_{\text{enclosed}} = 2.2125 \times 10^{-12} \text{ C}$

Rounding to one decimal place, the enclosed charge is $2.2 \times 10^{-12} \text{ C}$.