Answer
(a) $\Phi_{E} = 1.36 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$
(b) The same flux $\Phi_{E} = 1.36 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$
(c) $\Phi_{E} = 2.71 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$
Work Step by Step
(a) The electric field for the cylindrical shape is
$$ E=\frac{\lambda}{2 \pi \epsilon_{0} r} $$
and the area is $ A=2 \pi r l $. So, the electric flux through the cylinder is
\begin{align}
\Phi_{E}&=E A \cos \phi\\
& =E A \cos 0 \\
& =\left(\frac{\lambda}{2 \pi \epsilon_{0} r} \right)(2 \pi r l)\\
& =\frac{\lambda l}{\epsilon_{0}}\\
& =\frac{\left(3 \times 10^{-6} \mathrm{C} / \mathrm{m}\right)(0.4 \mathrm{m})}{8.854 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}}=\boxed{1.36 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}}
\end{align}
(b) As shown in part (a), the flux doesn't depend on the radius of the cylinder, so the flux is the same
$$ \boxed{\Phi_{E} = 1.36 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}}$$
(c) When the length of the cylinder becomes $l = 0.8$ m, the flux will be
\begin{align}
\Phi_{E}&=\frac{\lambda l}{\epsilon_{0}}\\
& =\frac{\left(3 \times 10^{-6} \mathrm{C} / \mathrm{m}\right)(0.8 \mathrm{m})}{8.854 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}}=\boxed{2.71 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}}
\end{align}
