I am having trouble in finding the remainder of $(1\cdot2\cdots102)^3\mod 105$
It is not possible to apply Wilson's Theorem here because 105 is composite. Can anybody help me?
$105 = 3 \cdot 5 \cdot 7, 102!$ is divisible all of $3,5$ and $7,$ therefore $102!^3$ is divisible $105$
Hint: Yes, $105$ is composite. So while you can't use Wilson's theorem, it actually makes the problem a lot easier, rather than more difficult. Think about exactly why Wilson's theorem fails for composite numbers: What is $(n-1)!\pmod n$ in those cases?