= Solution
The <phase averaging of a polarization Bell state> gives the ensemble <density matrix>
$$
\rho_{\varphi_0}=\frac12\left[|01\rangle\langle01|+|10\rangle\langle10|+s\bigl(|01\rangle\langle10|+|10\rangle\langle01|\bigr)\right],\qquad
s=\frac{\sin\varphi_0}{\varphi_0},
$$
with $s=1$ by continuity at $\varphi_0=0$. This follows from averaging $e^{i\varphi}$ over $[-\varphi_0,\varphi_0]$. An observable's <expectation> is linear in the <density matrix>, so averaging the pure-state Bell function is equivalent to evaluating the mixed-state <expectation>. Therefore
$$
\boxed{\mathcal B_{\varphi_0}=\sqrt2\left(1+\frac{\sin\varphi_0}{\varphi_0}\right).}
$$
Violation requires $\sin\varphi_0/\varphi_0>\sqrt2-1\simeq0.41421$. The supplied graph gives the first crossing near $2.1$ radians; solving more accurately gives
$$
\boxed{0\leq\varphi_0<2.09275\ldots\ \text{rad}.}
$$
The inequality is strict at the upper endpoint. On $(0,\pi)$ the sinc function decreases to zero, so this is its only crossing in that interval. For $\varphi_0\geq\pi$, $|\sin\varphi_0/\varphi_0|\leq1/\pi<\sqrt2-1$, excluding later violation windows. The plotted Bell signal shows the same threshold without copying the source illustration.
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