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I think it's strictly below $0'$. Namely let's call your number $\Gamma(\Omega)$ where $\Gamma$ is a Turing functional. Let $\Phi$ be any other Turing functional. Then show that the set $$ S = \{X: X = \Phi(\Gamma(X))\} $$ has measure 0 (which is easy since $\Gamma$ erases a lot of information about $X$) and moreover show that it is a Martin-Löf null set (this requires a bit more care). Then, since $\Omega$ is Martin-Löf random, it follows that $\Omega$ does not belong to $S$. Hence $\Omega$ is not Turing reducible to $\Gamma(\Omega)$.