Radiationless energy transfer: how to arrive at the equation that give R0 between two chromophores?
I think I found the answer to my own question through some help from some classmates. What the professor did, was to solve the equation mathematically by converting $R_0^6$ to $1$. 

So first, the efficiency of the transfer $E_\mathrm{eff}$ is actually a ratio of the actual transfer efficiency divided by the maximum efficient (which we should assume is "one", unless we are told otherwise. 

The next step is a bit hard to explain, as I am not that good in math, but I will try my best. What I understand is that you can "gather" the two $R_0^6$ by multiplying both the numerator and denominator by what we should think of as the number one, but we actually divide by $R_0^6$ and in this way we don’t have to include $R_0^6$ that we divide by on both sides of the equal sign - as it is technically the number one (mathematically allowed), as following: 
$$\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}} = \frac{\frac{R_0^6}{R_0^6}}{\frac{R_0^6}{R_0^6}+ \frac{r^6}{R_0^6}}$$

We then get as following:
$$\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}} = \frac{1}{1 + \frac{r^6}{R_0^6}}$$

So by applying some math and assumptions we can solve the equation like this, and what I have explained in my question should therefore be correct.