Is the vacuum permittivity a result of defining other units first? From Coulomb's law
$$F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$$
If instead we defined the charge of a coulomb to be $\sqrt{\epsilon_0}$ times it is now, the force would be exactly the multiplication of the charges divided by the surface area of the sphere of radius $r$ where $r$ is the distance from $q_1\;to\; q_2$. If instead the coulomb (or ampere) had been defined a different way, would this constant still exist? If so what is the origin of it and what does it mean? 
 A: The value of permittivity depends on the unit system you use. Pieter's reply shows its value in the SI (a.k.a. MKS - metre, kilo, second) system. In the electrostatic cgs system (centimetre, gram, second) its value is a dimensionless $1/4\pi$.
Wikipedia has more info.
A: Its value is just a consequence of the unit system. Nowadays, $\varepsilon_0$ is called the electric constant (of the SI units).
https://physics.nist.gov/cgi-bin/cuu/Value?ep0
A: 
If instead the coulomb (or ampere) had been defined a different way, would this constant still exist? 

Yes, but its meaning can change. In fact, we're in the middle of just such a redefinition, including a change in the status of the vacuum permittivity.


*

*In the current SI, the definition of the electrical-charge units via the ampere is equivalent to giving a fixed value to the vacuum permeability $\mu_0$. Since this can be combined with the vacuum permittivity to give the speed of light via $\varepsilon_0\mu_0=1/c^2$, this means that in the current SI the vacuum permittivity has an exact value,
$$
\varepsilon_0 = \frac{1}{c^2\mu_0} = \frac{1}{299792458^2\times 4\pi\times 10^{−7} \:\rm H\:m/s^2}.
$$

*In the new SI, on the other hand, the value of the coulomb will be fixed (via an explicit value of the elementary charge in coulombs), which means that the Coulomb constant $1/4\pi\varepsilon_0$ will acquire its old meaning - the (experimentally determined) force between two elementary charges at some fixed distance $r$.
In practice, that's almost certainly not the best way to measure it, of course: it is much better to re-express it as
$$
\varepsilon_0 = \frac{1}{\alpha} \frac{e^2}{2hc},
$$
where $e$, $h$ and $c$ have fixed values and $\alpha$, the fine-structure constant, is measured independently, so that 
(There is a comprehensive table of how the status of the different fundamental constants changes in the shift to the new system in the Wikipedia article.)
The question of what $\varepsilon_0$ "means" is a valid and interesting question, but it doesn't have a unique answer since the response depends on how you have defined the electric units. As described in e.g. this previous question, there are several valid approaches to defining electrical charge, even if you try to do away to do with units, and they will generally produce different results (involving one or more factors of the speed of light sneaking in in various places). The role of proportionality constants, such as the vacuum permittivity, is to grease those gaps and allow for a cleaner bridging of the different places where electrical units show up.
A: Yes, if the Coulomb's Law would have defined in a different way then definitely a constant should exist,it could have been any other constant instead of the vacuum permittivity but it would have exist. 
The introduction of the constant is due to the "Rationalization of units".
If we experimentally try to find the electrostatic force between two objects, surely the Coulomb's Law can only help us to find the very nearest value, not the exact.
Sometimes, in real situations there are lots of other factors such as air resistance, material quality,etc. that we have to take care of.
The introduction of a constant help us to keep a check on those factors and estimate a nearest result. 
