(b) For any integer $n$, we have $n|n \Rightarrow n\,R\,n$

So, $n\,R\,n$ for all $n \in Z \Rightarrow R$ is reflexive

Now $2|6$ but $6+2,==> (2,6)$$ \in R$ but $(6, 2)$ $\not \in R$

So, $R$ is not symmetric.

Let $(m,n) \in R$ and $(n,p) \in R$.

Then  $\left. {\begin{array}{*{20}{c}}
  {(m,n) \in R \Rightarrow m|n} \\ 
  {(n,p) \in R \Rightarrow n|p} 
\end{array}} \right] \Rightarrow m|p \Rightarrow (m,p) \in R$

So, $R$ is transitive.

Hence, $R$ is reflexive and transitive but it is not symmetric.