Electric flux through a spherical surface

[Boilermaker54]

Homework Statement

Three point charges are located near a spherical Gaussian surface of radius R = 6 cm. One charge (+3Q) is inside a sphere, and the others are a distance R/3 outside the surface.

What is the electric flux through this surface (Q = 6 C)

Homework Equations

I am aware of guass's law for a sphere. EA= Q(enclosed)/8.55e-12 A for sphere = 4Pi r^2

The Attempt at a Solution

I'm not sure what radius or Q charge should get put int ote equation, I'm still trying to understand the concepts behind these problems.

 

[Matterwave]

The problem tells you that the radius is 6cm and the charge inside the sphere is +3Q

 

[minimark1234]

ill try to provide some intuition behind this type of problem.

a flux is the amount of a volume passing a point per time. in this type of problem we can use electric field lines as our volume, and the point being the surface of the sphere. if one assumes that the field lines start at the center of the sphere, no field lines stop abruptly or ones begin at any other point, then no matter the surface area of the sphere the same number or field lines are passing.
Gauss's theroem says: \int(\nabla\cdot\vec{v})\:d\tau=\int\vec{v}\cdot d\vec{a}
since \int(\nabla\cdot\vec{E})\:d\tau=\frac{1}{\epsilon}q
\int\vec{E}\cdot d\vec{a}=\frac{1}{\epsilon}q

but since no matter what the change in area the same amount of electric field lines are passing through the surface area. the E is independent of the da, and can be taken out of the integral. giving u \vec{E}\int d\vec{a}=\frac{1}{\epsilon}q or \vec{E}A=\frac{1}{\epsilon}q A being 4\pi r^{2}
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that was my understanding on Gauss's theorem, but if i understood it incorrectly someone correct me before i confuse this good sir.