Given electric field

$\vec E = 4000x^2\,\hat i\;\text{V/m}$

So the field is along the $x$-direction and depends only on $x$.

Flux through the cube

Since $\vec E$ is along $\hat i$, only the two faces perpendicular to the $x$-axis contribute to net flux.

Using Gauss/divergence idea,

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

Now,

$\nabla\cdot \vec E = \frac{\partial}{\partial x}(4000x^2)=8000x$

From the figure / standard placement

The cube of side $20\text{ cm}=0.2\text{ m}$ is placed with its faces normal to axes, extending from

$x=0.1\text{ m} \text{ to } x=0.3\text{ m}$

(equivalently, left face at $10\text{ cm}$ and right face at $30\text{ cm}$).

Cross-sectional area:

$A=(0.2)^2=0.04\,\text{m}^2$

Flux by face method

At $x=0.3$ m:

$E_2=4000(0.3)^2=4000(0.09)=360\,\text{V/m}$

Outward flux through right face:

$\Phi_2=E_2A=360\times 0.04=14.4\,\text{V·m}$

At $x=0.1$ m:

$E_1=4000(0.1)^2=4000(0.01)=40\,\text{V/m}$

Outward normal on left face is along $-\hat i$, so flux there is negative:

$\Phi_1=-E_1A=-40\times 0.04=-1.6\,\text{V·m}$

Therefore net flux,

$\Phi=14.4-1.6=12.8\,\text{V·m}$

Convert into asked unit

Since

$1\,\text{V·m}=100\,\text{V·cm}$

we get

$\Phi=12.8\times 100=1280\,\text{V·cm}$

Final answer

$\boxed{1280}$

Comparison with stored answer

Stored correct answer is $640$, but the calculated flux is $1280$.

A value of $640$ would arise if one mistakenly used half the required net flux. For the cube spanning $x=10\text{ cm}$ to $30\text{ cm}$, the correct net flux is clearly $1280\,\text{V·cm}$.