A construction of $M \models \neg B\Sigma_2 + (1)$:

Firstly, let $N$ be a countable non-standard model of $PA$. We build $M$ as $N[G]$ by forcing.

Fix some non-standard $b \in N$. The poset consists of pairs $(f,c,d)$, where $f$ is an $N$-finite partial function with domain contained by $b \times N$, $0 < c < d < b$ and $d - c > \omega$.

$(f',c',d') \leq (f,c,d)$ iff

1. $f'$ extends $f$, $c \leq c' < d' \leq d$,
2. $f'(x,s) = 0$ for $(x,s) \in \operatorname{dom} f' - \operatorname{dom} f$ and $c \leq x < c'$,
3. $f'(y,s) = 1$ for $(y,s) \in \operatorname{dom} f' - \operatorname{dom} f$ and $d' < y \leq d$.

For each $(f,c,d)$ and $e$, we can extend $(f,c,d)$ to $(f',c,d)$ forcing one of the followings:

(a) $\Phi_e(G)$ is partial;

(b) $\Phi_e(f'; n) \downarrow = \sigma$ and $\sigma$ is the $\prec$-least $\tau$ s.t. there are $g, m$ with $(g,c,d) \leq (f,c,d)$ and $\Phi_e(g; m) \downarrow = \tau$.

Of course, we can achieve (b) because $N \models PA + (1)$. For sufficiently generic $G$, $N[G]$ contains a $\Delta_2$ cut $I < b$. By (b), $N[G] \models (1)$.

However, does $B\Sigma_2 + (1)$ imply $I\Sigma_2$?

A construction of $M \models \neg B\Sigma_2 + (1)$:

Firstly, let $N$ be a countable non-standard model of $PA$. We build $M$ as $N[G]$ by forcing.

Fix some non-standard $b \in N$. The poset consists of triples $(f,c,d)$, where $f$ is an $N$-finite partial function with domain contained by $b \times N$, $0 < c < d < b$ and $d - c > \omega$.

$(f',c',d') \leq (f,c,d)$ iff

1. $f'$ extends $f$, $c \leq c' < d' \leq d$,
2. $f'(x,s) = 0$ for $(x,s) \in \operatorname{dom} f' - \operatorname{dom} f$ and $c \leq x < c'$,
3. $f'(y,s) = 1$ for $(y,s) \in \operatorname{dom} f' - \operatorname{dom} f$ and $d' < y \leq d$.

For each $(f,c,d)$ and $e$, we can extend $(f,c,d)$ to $(f',c,d)$ forcing one of the followings:

(a) $\Phi_e(G)$ is partial;

(b) $\Phi_e(f'; n) \downarrow = \sigma$ and $\sigma$ is the $\prec$-least $\tau$ s.t. there are $g, m$ with $(g,c,d) \leq (f,c,d)$ and $\Phi_e(g; m) \downarrow = \tau$.

Of course, we can achieve (b) because $N \models PA + (1)$. For sufficiently generic $G$, $N[G]$ contains a $\Delta_2$ cut $I < b$. By (b), $N[G] \models (1)$.

However, does $B\Sigma_2 + (1)$ imply $I\Sigma_2$?