= Solution
Take a normalized environment <quantum state>
$$
\boxed{|\psi_E\rangle=\sqrt{1-\epsilon}|0\rangle+\sqrt\epsilon|1\rangle.}
$$
The <Controlled-Z gate> is symmetric between the two <qubits>, so it can be written with the environment as the controlling system:
$$
C_Z=I\otimes|0\rangle\langle0|+Z\otimes|1\rangle\langle1|.
$$
Expanding the joint <density operator> after the interaction gives
$$
\begin{aligned}
C_Z(\rho\otimes|\psi_E\rangle\langle\psi_E|)C_Z
={}&(1-\epsilon)\rho\otimes|0\rangle\langle0|+\epsilon Z\rho Z\otimes|1\rangle\langle1|\\
&+\sqrt{\epsilon(1-\epsilon)}\bigl(\rho Z\otimes|0\rangle\langle1|+Z\rho\otimes|1\rangle\langle0|\bigr).
\end{aligned}
$$
In the <partial trace>, the two off-diagonal environment operators have trace zero, while each diagonal projector has trace one. Hence
$$
\boxed{\operatorname{tr}_E\!\left[C_Z(\rho\otimes|\psi_E\rangle\langle\psi_E|)C_Z\right]
=(1-\epsilon)\rho+\epsilon Z\rho Z=D_\epsilon(\rho).}
$$
Equivalently, the <Kraus operators> are $K_0=\sqrt{1-\epsilon}I$ and $K_1=\sqrt\epsilon Z$, and $K_0^\dagger K_0+K_1^\dagger K_1=I$. This explicit <Stinespring dilation> describes the system's dephasing by discarding information in the environment. A fresh environment in this state at each storage interval also realizes the memoryless iteration used in part (a).
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