Consider the $ABC$ conjecture in the following form:
For every positive real number $\epsilon$, there exists a constant $k(\epsilon)$ such that for all triples $(a, b, c)$ of coprime positive integers, with $a + b = c$:
$c<k(\epsilon)\cdot rad(abc)^{1+\epsilon}$
Suppose I prove for a particular choice of $k(\epsilon)$ and any choice of $\epsilon>0$ that:
$c \ge k(\epsilon)\cdot rad(abc)^{1+\epsilon}$
What is the implication for the $ABC$ conjecture?
I'm not certain that I understand exactly what it is you are asking, but there are definitely interesting applications to knowing the ABC conjecture for particular values of $\epsilon, k(\epsilon)$.
To illustrate one such application, suppose that the ABC conjecture holds with $\epsilon=k(\epsilon)=1$. Then for all triples $(a,b,c)$ of relatively prime integers with $a+b=c$ we would have $$c\leq \left(\prod_{p\mid abc} p\right)^2.$$
Now suppose that $x,y,z$ are positive, relatively prime integers such that $x^n+y^n=z^n$, where $n$ is a positive integer. If we let $a=x^n, b=y^n, c=z^n$ then $$z^n=c\leq \mathrm{rad}(abc)^2 = \mathrm{rad}(xyz)^2 \leq (xyz)^2 < z^6.$$ Thus we've shown that $n<6$ and consequently that Fermat's Last Theorem holds for all $n\geq 6$.