There's not enough information in the question. If $M\models V=L$, then in $M$ we have that $\mathrm{HOD}^M=L^M=M$, and therefore $\mathrm{HOD}^{\mathrm{HOD}^M}=L^M=M$ as well.
In particular we have that in both instances sets of real definable from ordinals and reals (in $M$, which is the same as $\mathrm{HOD}^M$ in this case) are the same sets of reals in $L$, and so there are sets without the Baire property.