= Solution
Let the original cylinder axis be $\mathbf r=\mathbf0$, and let the hole axis be at transverse <vector> $\mathbf a$, with $|\mathbf a|=a$. The <charge density> is reproduced by a full cylinder of <radius> $R_1$ and <charge density> $+\rho$, superposed with a cylinder of <radius> $R_2$ centred at $\mathbf a$ and <charge density> $-\rho$. This uses the <superposition principle> for fixed insulating <electric charges>; the hole is not a hole in an <electrical conductor>.
For a complete uniformly charged infinite cylinder of <radius> $R$, a coaxial <Gaussian surface> of <length> $\ell$ gives, by <Gauss's law>,
$$
2\pi s\ell E(s)=\begin{cases}\rho\pi s^2\ell/\epsilon_0,&s<R,\\
\rho\pi R^2\ell/\epsilon_0,&s\geq R.
\end{cases}
$$
Consequently define
$$
\mathbf F_R(\mathbf s)=\begin{cases}\mathbf s,&|\mathbf s|\leq R,\\
R^2\mathbf s/|\mathbf s|^2,&|\mathbf s|>R.
\end{cases}
$$
The complete <electric field>, including the exterior if desired, is
$$
\boxed{\mathbf E(\mathbf r)=\frac\rho{2\epsilon_0}
\left[\mathbf F_{R_1}(\mathbf r)-\mathbf F_{R_2}(\mathbf r-\mathbf a)\right].}
$$
In the insulating material, $|\mathbf r|<R_1$ and $s=|\mathbf r-\mathbf a|>R_2$, so
$$
\boxed{\mathbf E_{\rm material}(\mathbf r)=\frac\rho{2\epsilon_0}
\left[\mathbf r-\frac{R_2^2(\mathbf r-\mathbf a)}{|\mathbf r-\mathbf a|^2}\right].}
$$
Inside the hole both contributions use their interior formulas, and all dependence on $\mathbf r$ cancels:
$$
\boxed{\mathbf E_{\rm hole}=\frac\rho{2\epsilon_0}[\mathbf r-(\mathbf r-\mathbf a)]
=\frac\rho{2\epsilon_0}\mathbf a.}
$$
The <electric field> in the hole is uniform, points from the original axis towards the hole axis for $\rho>0$, and vanishes for a concentric hole. The two expressions agree at the hole wall, as expected with no imposed <surface charge density> there. Its uniformity is the useful <uniform electric field in an off-axis cylindrical cavity> result.
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