There are $B$ with this property.

Lets first recall some definitions and notation. Suppose $X \in 2^\omega$ (which we identify with subsets of $\N$ via characteristic functions). Then $X'$ is the Turing jump of $X$, and $X^{(n)}$ is the nth iterate of the Turing jump of $X$. $X$ is said to be $n$-generic (i.e. $\Sigma^0_n$-generic for Cohen forcing) if for every $\Sigma^0_n$ subset $S$ of $2^{< \omega}$ there is a finite initial segment $\sigma$ of $X$ so that either $\sigma \in S$, or there are no extensions of $\sigma$ contained in $S$. $X$ is said to be arithmetically generic if $X$ is $n$-generic for every $n$. It is a standard fact that if $X$ is $1$-generic, then $X' \equiv_T 0' \oplus X$, where $\oplus$ is recursive join. We also have that if $X$ is $n$-generic and $Y$ is $1$-generic relative to $X \oplus 0^{(n-1)}$, then $X \oplus Y$ is $n$-generic. Finally, if $X_0, X_1, \ldots \in 2^\omega$, then $\bigoplus_n X_n = \{\langle n, m \rangle: m \in X_n\}$ notes their recursive join.

Let $A = 0^{(\omega)} = \bigoplus_{n} 0^{(n)}$; the set of true sentences in first order arithmetic. Below, we will construct a $B$ so that:

- If $n \notin A$, then $B_n$ is arithmetically generic, and if $n \in A$, then $B_n$ is $(n+1)$-generic and computable from $0^{(n+1)}$. Hence $\{n \in \mathbb{N} : \text{$B_n$ is arithmetic}\}$ is equal to $A$.
- For each $k \in \N$, let $m_i$ be the $i$th element of the set $\{m \in \N : m \notin A \lor m \geq k\}$. Then $C_k = \bigoplus_{i \in \omega} B_{m_i}$ is $(k+1)$-generic.

A $B$ with the above two properties gives a positive answer to your question by the following reasoning. We can prove by induction that for any $n$, $B^{(n)} \equiv_T 0^{(n)} \oplus C_{n}$. This is trivial when $n = 0$. Now for the inductive case, assume $B^{(n)} \equiv_T 0^{(n)} \oplus C_{n}$. Then we see that $B^{(n+1)} \equiv_T (0^{(n)} \oplus C_n)' \equiv_T 0^{(n+1)} \oplus C_n$ since $C_n$ is $(n+1)$-generic and hence $1$-generic relative to $0^{(n)}$. Finally, $0^{(n+1)} \oplus C_n \equiv_T 0^{(n+1)} \oplus C_{n+1}$ because either $n \notin A$ and so $C_n = C_{n+1}$ or $n \in A$ so $C_{n+1} \equiv_T B_n \oplus C_n$, but then $B_n \leq_T 0^{(n+1)}$.

Hence, $B^{(n)}$ cannot compute $A$ for any $n \in \omega$, since $C_{n}$ is $(n+1)$-generic, and so $0^{(n)} \oplus C_{n} \ngeq_T 0^{(n+1)}$. (Indeed, it's easy to see that $B$ is $GL_n$ for every $n \in \N$). Thus, $A$ is not arithmetically definable relative to $B$.

So lets turn now to the construction of a $B$ with the required properties (1) and (2) above. We construct $B$ in countably many steps where after step $n$ we will have completely defined $B_0, B_1, \ldots, B_n$ and finitely many other bits of $B$ (i.e. finitely many bits of finitely many $B_i$ where $i > n$). For step $0$, let $B_0$ be an arbitrary real satisfying condition (1).

Given $k \leq n$, let $m_0, \ldots, m_j$ be the elements of $\{m \in \mathbb{N}: m \notin A \lor m \geq k\} \cap \{0, \ldots, n\}$, and define $C_{k,n} = B_{m_0} \oplus \ldots \oplus B_{m_j}$. After each stage $n$, we will have ensured that $C_{n,k}$ is $(k+1)$-generic for every $k \leq n$.

Now at step $n > 0$, for each pair $(i,k)$ where $i,k < n$, do the following. Let $S_{i,k}$ be the $i$th $\Sigma^0_{k+1}$ subset of $2^{< \omega}$. If we can find a finite extension of our approximation of $B$ so that the resulting approximation of $C_k$ extends an element of $S_{i,k}$, then extend our approximation of $B$ in this way. If this is not possible, then since $C_{k,n-1}$ is $(k+1)$-generic, there must be some finite subset of our current approximation to $B$ which cannot be extended to extend an element of $S_{i,k}$.

Next, we finish step $n$ by defining $B_n$. If $n \notin A$ pick $B_n$ to be an element of $2^{\omega}$ extending the finite approximation of $B_n$ that we currently have, which is arithmetically generic relative to $B_0 \oplus \ldots \oplus B_{n-1}$. This clearly ensures that $C_{k,n}$ will be $(k+1)$-generic for each $k < n$, since $C_{k,n-1}$ is $(k+1)$-generic, and $B_n$ is $(k+1)$-generic relative to it. Similarly, $C_{n,n}$ is clearly $(n+1)$-generic. Otherwise, if $n \in A$, let $B_n$ be an arbitrary $(n+1)$-generic computable from $0^{(n+1)}$ and extending the finitely many bits of $B_n$ we have already determined. Now for any $k < n$, let $j_0, \ldots, j_t$ be the elements of $A$ that are $\geq k$ and $< n$. Then since $B_n$ is $1$-generic relative to $0^{(n)}$ which can compute $B_{j_0} \oplus \ldots \oplus B_{j_t}$ we have that $B_{j_0} \oplus \ldots \oplus B_{j_t} \oplus B_n$ is $(k+1)$-generic. Hence $C_{k,n}$ is $(k+1)$-generic, since the remaining elements in the finite join defining $C_{k,n}$ are mutually arithmetically generic (and are hence $(k+1)$-generic relative to $B_{j_0} \oplus \ldots \oplus B_{j_t} \oplus B_n$). Similarly, $C_{n,n}$ is again clearly $n+1$-generic.

To verify that our construction works, note that condition (1) is satisfied by how we pick $B_n$ in the last paragraph. Condition (2) is satisfied by the paragraph before that.