Suppose that we have two different numbers, $A, B$, and arbitrary numbers $c,d$. Of course, they're all integers.
If we've found $\gcd(A,c)$ by using Euclidean Algorithm at $n$th iteration. Whereas, it takes $n-1$ times of iteration to find $\gcd(B,d)$. Can we say that $A$ is larger than $B$?
If not, can we make some modifications to the statement so to make it true?
No, you can't, and you can't even if $c=d$ and you know this for several different values of $c$.
For example, if you have an arbitrary list $c_1,c_2,\ldots,c_k$ of integers, all at least $3$, consider computing $\gcd(A,c_i)$ and $\gcd(B,c_i)$, where $A=2\prod_{i=1}^rc_i-1$ and $B=n\prod_{i=1}^rc_i+1$ for any $n\geq 2$.
For each $i$ it will take two steps to compute $\gcd(B,c_i)$ (since the first remainder is $1$ which equals the GCD) but three to compute $\gcd(A,c_i)$ (since the first remainder is $c_i-1$ and the second remainder is $1$), and $B>A$.
The principle here is that if you have a finite list of values $c_i$, and you know that $\gcd(A,c_i)$ takes $\ell_i$ steps for each $i$, then there are infinitely many other values $A'$ with the same property. So these lengths $\ell_i$ tell you essentially nothing about the size of $A$.