For reflexivity for all $x \in I, R(x,x)$ but this is violated for $x = 0$. So, given relation is not reflexive.
For symmetry, for all $x,y \in I, R(x,y) \implies R(y,x).$ This is violated for $x = 2, y = 1$.
For anti-symmetry, for all $x,y \in I, (R(x,y) \wedge R(y,x)) \implies x = y$. We have $R(-1, 1)$ and $R(1,-1)$ so $R$ is not anti-symmetric.
For transitivity, for all $x,y,z \in I, R(x,y) \wedge R(y,z) \implies R(x,z).$ As per the rule of division this holds.
So, option B is the answer.