Electric field flux through the cubic with surface density of charge first of all I want you to excuse me for my level of English.
I'm doing a homework on Gauss's law, specifically on how to calculate electric flux through a cubic surface. Below I put the complete statement :

We have a charge distribution formed by a point charge $q = 13 \rm nC$ at the origin of coordinates and a surface density of charge $\sigma = −29 \rm pC/ cm^2$ , flat, infinite, parallel to the $xz-$plane in the coordinate $y = a$, with $a = 7 \rm cm$. We also have a closed surface of cubic form of side $\mathrm L$, centered in the origin.
a) Calculate the value of the electric field flux through the cubic surface with side $\rm L=19 cm$.


What I know is that according to Gauss's law, we can calculate the flow knowing the charge enclosed by the cube, so Ill be able to resolve this if we had no surface charge density using

\begin{equation}
  \Phi=\iint\limits_{\texttt{cube}}\mathbf E\boldsymbol{\cdot}\mathrm d\mathbf S=\dfrac{Q}{\varepsilon_0}
\tag{01}\label{01}  
\end{equation}

But my question here is, how does this affect to charge inside?
Does this affect at all, if so, how do I calculate the electric field flux through the cubic surface?
 A: Hint: The flux through the surface has two contributions that can be added: 1. The flux due to the uniform electric field of the infinite sheet of charge and 2. The flux due to the electric field from the point charge at the center.
Hope this helps.
A: you can calculate the flux through the surface of the cube, by finding the electric field then calculating the flux through the face of the cube.
First, calculate the electric field
the electric field for a point charge is
$$\underline{E}=\frac{Q}{(4\pi\epsilon_0)(\sqrt{x^2+y^2+z^2})^3}*(x\widehat{\underline{i}}+y\widehat{\underline{j}}+z\widehat{\underline{k}})$$
and the electric field for a sheet a of charge in the $x$-$z$ plane at $y=a$, is
$$
\underline{E}=
\begin{cases}
 -\frac{\sigma}{2\epsilon_0}\widehat{\underline{j}}&\text{if}\, y<a\\
 +\frac{\sigma}{2\epsilon_0}\widehat{\underline{j}}&\text{if}\, y>a\\
      0\widehat{\underline{j}}&\text{if}\,y=a
\end{cases}
$$
so the total electric field would be the sum of these two fields.
Second, calculate the flux
the electric flux is give by
$$E_{flux}=\int\int _R \underline{E}\cdot dA$$
For this case this becomes, withe the cube face being $c\ge x\ge -c$ and $c\ge y\ge -c$ with $c=\frac{19}{2}$
$$E_{flux}=\int_{-c}^c\int_{-c}^c  (\underline{E}\cdot \widehat{\underline{k}}) dxdy$$
therefore $\underline{E}\cdot \widehat{\underline{k}}$ is
$$\underline{E}\cdot \widehat{\underline{k}}=\frac{Qz}{(4\pi\epsilon_0)(\sqrt{x^2+y^2+z^2})^3}$$
so the flux is through the face of the cube at $z=\frac{19}{2}$
$$E_{flux}=\int_{-c}^c\int_{-c}^c  (\frac{Qz}{(4\pi\epsilon_0)(\sqrt{x^2+y^2+z^2})^3}) dxdy$$
which is the same flux calculation as is the sheet of charge wasn't there, which we can calculate without using integrals.
calculating the flux from a point charge in the centre of a cube
The total flux from the point charge is $\frac{q}{\epsilon_0}$, and this is split through the six faces. Therefore the flux through a face is $\frac{q}{6\epsilon_0}$.
So in your question the flux through the face of the cube is $\frac{q}{6\epsilon_0}$
hopefully that helps
