Given electric field

$\vec E = 2x\,\hat i \quad (\text{N/C})$

So the field has only an $x$-component:

$E_x = 2x$

Use Gauss's divergence theorem

Net electric flux through a closed surface is

$\Phi = \iiint (\nabla \cdot \vec E)\, dV$

First compute divergence:

$\nabla \cdot \vec E = \frac{\partial (2x)}{\partial x} = 2$

Volume of the cube

Side of cube $=2\,\text{m}$

$V = 2^3 = 8\,\text{m}^3$

Flux through the cube

Since divergence is constant,

$\Phi = (\nabla \cdot \vec E)\,V = 2 \times 8 = 16$

Therefore,

$\boxed{\Phi = 16\,\text{N m}^2/\text{C}}$

Check with stored answer

Stored correct answer = $16$

This matches our derived result.