The question says:- $A = \{1,2,3,4,5,6\}$ and $R = \{(S_1, S_2) :S_1, S_2 \subset A, S1\nsubseteq S2\}$.
My thought:- $S_1$ contains the subsets of $A$ and $S_2$ contains the subset of $A$ and $S_1$ is not a subset of $S_2$ then there are no elements common in $S_1$ and $S_2$ so there is not relations such as if $(a, b)$ is an element of $A$ and $(b, c)$ is an element of $A$ then $(a, c)$ is an element of $A$ as $b$ is not element of $S_1$.. Hence i can say the relation is transitive.. But the answer says it is not transitive
Assuming I interpreted your relation correctly, as I described in the comment...
For transitivity to hold, we require $\forall S_1, S_2, S_3$, subsets of $A$, $S_1 \sim S_2$ and $S_2 \sim S_3 \implies S_1 \sim S_3$. (or, $(S_1, S_2) \in R$ and $(S_2, S_3) \in R \implies (S_1, S_3) \in R$)
Let $S_1 = \{1, 2\}$
Let $S_2 = \{2, 3\}$
Let $S_3 = \{1, 2, 4\}$
$S_1 \sim S_2$ (i.e. $(S_1, S_2) \in R$) since $S_1$ is not a subset of $S_2$.
$S_2 \sim S_3$ (i.e. $(S_2, S_3) \in R$) since $S_2$ is not a subset of $S_3$.
But $S_1$ is a subset of $S_3$, so $S_1$ is not equivalent to $S_3$ ($(S_1, S_3) \notin R$). This is one of many examples where transitivity will fail.