Using Fermat's Little Theorem to find the smallest k for which $a ^k $≡ 1(mod 11) for a = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. We know that $a^{11-1} ≡1\bmod{11}$ for all positive integers $a$ less than $11$, by Fermat’s Little
Theorem.
But $(11 − 1)$= $10$ is not the smallest k for which $a ^k$ ≡ 1(mod 11) for all such a.How do we find the smallest k for which $a ^k $≡ $1(mod 11)$ for a = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
$[$Examples: (i) $3^5$ ≡ 1(mod 11). Here k = 5 for a = 3.
(ii)$10^2$ ≡ $1(mod 11)$. Here k = 2 for a =10.$]$
How are the values of k related to $(11 − 1)= 10?$

You are seeking the Carmichael lambda function. The Wikipedia article shows how to compute this function using Euler's totient function and the lcm function.
EDIT: if you mean that the $a$ are independent of each other, then you are asking for a computation of multiplicative order, which is known to be a difficult problem. As far as I am aware, there is no direct formula for it.
