Answer
a) $Q_{enc} = -5.9 \times 10^{10} C $
b) within and outside.
Work Step by Step
a)
$\Phi$
$$
\begin{aligned}
& \Phi_{1}=\vec{E} \cdot \vec{A}_{1}=\left|E_{1} \| A_{1}\right| \cos \phi_{1} \\
& \Phi_{1}=\left(2.5 \times 10^{4} \mathrm{~N} / \mathrm{c}\right)(0.06 \mathrm{~m})(0.05 \mathrm{~m}) \cos 60^{\circ} \\
& \Phi_{1}=37.5 \frac{\mathrm{N}}{\mathrm{c}} \cdot \mathrm{m}^{2} \quad(\text { saliante) }
\end{aligned}
$$
$\Phi_{2}$
$$
\begin{aligned}
& \Phi_{2}=\vec{E}_{2} \cdot \vec{A}_{2}=E_{2} A_{2} \cos \phi_{2} \\
& \Phi_{2}=\left(7.0 \times 10^{4} \mathrm{~N} / \mathrm{c}\right)(0.06 \mathrm{~m})(0.05 \mathrm{~m}) \cos 60^{\circ} \\
& \Phi_{2}=-105 \frac{\mathrm{N}}{\mathrm{c}} \cdot \mathrm{m}^{2} \text { (entrante) }
\end{aligned}
$$
$\Phi_{\text {Neto }}$
$$
\begin{aligned}
& \Phi_{\text {Neto }}=\Phi_{1}+\Phi_{2}=37.5 \frac{\mathrm{N}}{\mathrm{c}} \cdot \mathrm{m}^{2}-105 \frac{\mathrm{N}}{\mathrm{C}} \cdot \mathrm{m}^{2}=-67.5 \frac{\mathrm{N}}{\mathrm{C}} \cdot \mathrm{m}^{2} \\
& \Phi_{\text {Neto }}=\frac{Q_{\text {enc }}}{\varepsilon_{0}} \Rightarrow Q_{\text {enc }}=\Phi_{\text {Neto }} \varepsilon_{0} \\
& Q_{\text {enc }}=\left(-67.5 \frac{\mathrm{N}}{\mathrm{c}} \mathrm{m}^{2}\right)\left(8.85 \times 10^{-12} \cdot \frac{\mathrm{C}^{2}}{\mu_{0}}\right) \\
& Q_{\text {enc }}=-5.9 \times 10^{-10} \mathrm{C}=-0.59 \mathrm{n} \mathrm{C}
\end{aligned}
$$
b) If there were no charge inside the parallelepiped, the net flow would be zero. This is not the case, so there is charge inside. The electric field lines that pass through the surface of the parallelepiped must end in charges, so there must also be charges outside the parallelepiped.
