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**Hint:**The region around a charged particle where other charged particles experience a force is known as the electric field. We can find the magnitude of electric field intensity from the electric field intensity vector(E) and then by substituting the value in the formula, the net flux will be obtained.

**Formula Used:**

Area of the square (A) = ${({\text{Side}})^2}$

Flux $(\Phi ) = \left| {\overrightarrow {\text{E}} } \right| \times {\text{A}} \times \cos \theta $

**Complete step by step solution:**

Electric field intensity (E) = ${\text{3}} \times {\text{1}}{{\text{0}}^3}{{\hat iN}}{{\text{C}}^{ - 1}}$

Magnitude of electric field intensity \[ = \left| {\overrightarrow {\text{E}} } \right| = \left| {{\text{3}} \times {\text{1}}{{\text{0}}^3}\hat i} \right| = \sqrt {{{({\text{3}} \times {\text{1}}{{\text{0}}^3})}^2}} = {\text{3}} \times {\text{1}}{{\text{0}}^3}{\text{ N/C}}\]

Side of the cube (s) = $20{\text{cm = 0}}{\text{.2m}}$ [$1{\text{m = 100cm}}$]

The electric field lines are passing in the x-direction, so the lines will pass through one of the sides of the cube and leaves through the opposite side of the cube. To obtain net flux we should calculate the flux through these sides of the cube which are squares.

We can calculate the area of the square by using the formula

Area of the square (A) = ${({\text{Side}})^2}$

$ \Rightarrow {\text{A}} = {\text{ (0}}{\text{.2}}{{\text{)}}^2} = {\text{ 0}}{\text{.04}}{{\text{m}}^2}$

Net flux can be calculated by using the formula

Flux $(\Phi ) = \left| {\overrightarrow {\text{E}} } \right| \times {\text{A}} \times \cos \theta $

As the sides of the cube are parallel to the coordinate axis, the angle between the electric field lines and normal to the surface will be $0^\circ $

Now, by substituting the values of electric field intensity, area of the square and $\theta $ in the above formula, we get

$ \Rightarrow \Phi = 3 \times {10^3} \times 0.04 \times {\text{cos(0)}}^\circ $

$ \Rightarrow \Phi = 0.12 \times {10^3} \times 1$

On further calculation, we get

$ \Rightarrow \Phi = 120{\text{N}}{{\text{m}}^2}{\text{/C}}$

**Therefore, The net flux through the cube is $120{\text{N}}{{\text{m}}^2}{\text{/C}}.$**

**Note:**While doing the calculation, all the quantities should be in the same unit. The given value of the side of the cube is in centimetres, so convert it into meters before calculating the area of the square.

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