My solution for this question is:

Reduction from $L_A$ to $A$, in the following way $f(x)=langle M_xrangle$

Emphasis: $exists$ 10 different words $w_1 ,dots,w_{10}in A$, otherwise $A$ finite $Rightarrow Ain R Rightarrow Ain RE$ in contradiction to $Anotin RE$

$M_x$ on input $y$ implemented in the following way:

- $M_x$ accept 9 of 10 words $w_1 ,dots,w_{10}$ that different from $y$ (because all the words different at least 9 words that different from $y$ exists)
- if $y$ equals to $w_i$, $M_x$ accept

The reduction defined for all imputes, can be realized, validity:

$yin ARightarrow M_x$ accepts 10 words from $ARightarrow langle M_x rangle in L_A$

$ynotin ARightarrow M_x$ accepts 9 words from $ARightarrow langle M_x rangle notin L_A$

$yin ALeftrightarrow langle M_x rangle in L_A$ it means $Ale L_A$, given $Anotin RE$ so $L_A notin RE$

Did I make a mistake somewhere?