Assumption: Encoding of Turing Machine M and input string w <M,w> is done using the alphabets which belong to $\Sigma$
Consider a computable function $f:\Sigma^* \to \Sigma^*$ defined as follows
$$f(x) = \{x^R | x \ \epsilon \ \Sigma^* \& \ x^R is \ string \ reverse \ operation \}$$
Certainly if $x \ \epsilon \ \Sigma^*$ then $f(x) \ \epsilon \ \Sigma^* $.
Now, for all valid encodings of the form $<M,w>$ which belong to language A, it is evident that $<M,w>^R$ belong to $\Sigma^*$. Therefore
$$<M,w> \epsilon \ A \to f(<M,w>) \ \epsilon \ B$$
Thus A is mapping reducible to B
This reasoning I gave was wrong, and I thank @Habibkhan for pointing that out. Here I'm still keeping this answer for other as a reference to avoid such mistakes.
Here is the mistake in my reasoning. The condition of mapping reducible is biconditional. So yes, if $<M,w> \epsilon \ A$ is true the RHS will certainly be true. But the implication other way round will not hold true, always.
However, this still cannot be used as a reason to say that $A \nless_m B$. The reasoning offered by @Habibkhan in his answer is in a sense the most appropriate way of doing it.
