What defines the strength of a Higgs Field? How can the strength of a Higgs Boson Field be measured?It is a scalar field and doesn't have a source unlike other fields such as gravity.What would the units be for a strength of the field at a point? The strength I'm looking for is a relationship between particle mass and it's interaction with the Higgs field.Something like $1g/1N$ (Particle collides with Higgs field with $1N$ of force and thus has a mass of $1$ gram).I know it's not that simple but how can a particles mass be related to it's interaction with the Higgs Boson Field?
 A: The useful way to think of mass here is through $E=m_0c^2$: it is the energy a particle has just by existing, before you add kinetic and potential energies.  (The more usual property of mass in terms of inertia follows.)
The $W$ boson has no 'intrinsic' mass so if there were no Higgs field present you would have $M_W=0$. But because we live in a universe with broken symmetry there is this residual Higgs field everywhere: let's call the value $v$; a $W$ interacts with this omnipresent Higgsness and this requires an energy ${1 \over 2} g_2 v$. (See e.g. Kane: Modern Elementary Particle Physics, p108). It makes sense that this is proportional to $v$, $g_2$ is the strength of the (weak) coupling, and the 2 just comes out of the algebra.   So we have (letting $c=1$ as usual)  $M_W={g_2 v \over 2}$.
The strength of the weak interaction in $\beta$ decay, the Fermi constant $G_F$ is well measured by nuclear physicists to be $1.17 \times 10^{-5} GeV^{-2}$.  It depends on the basic weak couplin $g_2$ and on $M_W$, by ${G_F \over \sqrt 2}={g_2^2 \over 8 M_W^2}$.
Putting these together gives $v={1 \over \sqrt{\sqrt 2 G_F}}$, and putting in the Fermi constant value gives $v=246$ GeV.  
So the residual Higgs field was known - accurately- well before the discovery of the Higgs particle and the measurement of its mass.  It was quite annoying. 
The $Z$ boson also acquires a mass in this way but is complicated by getting mixed up with the electromagnetic coupling. Quarks and leptons also get masses in the same way, but we don't know their coupling strength to the field apart from through their masses, so you can't use them to find $v$. But the $W$ gives it directly, because it has a gauge coupling $g_2$ which is connected to other quantities, rather than an arbitrary Yukawa coupling.  
A: The usual units for the strength of the Higgs field are gigaelectronvolts. The vacuum expectation value of the field (at every point!) is 246 GeV. Particle physicists generally use natural units where $\hbar$ and $c$ are $1$; mass, energy, and momentum are measured in GeV; and length and time are measured in GeV$^{-1}$.
The self-interacting Higgs field $H$ has a potential energy density $V$ (the famous “Mexican hat” potential which is the logo of Physics StackExchange) containing a term proportional to $H^4$, multiplied by a dimensionless constant. Since energy densities are measured in GeV$^4$, $H$ is measured in GeV. 
Yukawa interactions between the Higgs field and the fields of quarks and charged leptons give mass to those fermions when the Higgs acquires its nonzero vacuum expectation value. The Yukawa coupling constants are small dimensionless numbers measuring how strongly the fields interact. For example, for the electron to get its mass of 511 keV, the Yukawa coupling between the Higgs field and the electron-positron field would be on the order of this ratio, 0.000002. (The details are somewhat more complicated.) 
