If $B$, $\overline{B}\neq \varnothing$ , then for every recursive set $A$, $A \leq_m B$
This should be a comment on the original post by @Drupalist and on Ds D's answer, but it's too long. It's not enough to say that $A\le_M B$ requires $B$ be recursive or that one of $B, \overline B$ be r.e.

Let $A$ be recursive and $B=\{(\langle\,M\,\rangle,\langle\,N\,\rangle)\mid M,N\text{ are TMs and }L(M)=L(N)\}$. This language (sometimes called $\text{EQ}_{\text{TM}}$) is known to be neither r.e nor co-r.e. However, let

      M(x) =                    N(x) =
         return accept             if x = 0
                                      return reject
                                   else
                                      return accept


We have $(\langle\,M\,\rangle,\langle\,M\,\rangle)\in B$ and $(\langle\,M\,\rangle,\langle\,N\,\rangle)\in \overline{B}$. Then if $A$ is recursive, there is a decider TM, $D$, for A and if we define $f$ by

      f(x) =
         run D on x
         if D(x) = accept
            return (&lt;M&gt;, &lt;M&gt;)
         else
            return (&lt;M&gt;, &lt;N&gt;)


Then $f$ is Turing-computable and $x\in A\Longleftrightarrow f(x)\in B$, which is the definition of $A\le_M B$. So we can have such a reduction in every case where it's possible to find (not necessarily by a TM) two instances $y\in B$ and $n\in\overline{B}$.

Note that I'm not claiming that this technique will work for all possible $B$ nor am I claiming that this technique is the only possible one. All I've shown is that the problem isn't as simple as it seems at first glance.