Answer
(a) $E = 6.50 \times 10^{5} \mathrm{N} / \mathrm{C}$
(b) $E = 7.20 \times 10^{4} \mathrm{N} / \mathrm{C}$
Work Step by Step
(a) At y = 0.2 m, both lines apply electric field in the positive direction of y-axis, so the net electric field due to both lines equals the summation of their electric field
\begin{align}
E&=\frac{1}{2 \pi \epsilon_{0}} \frac{\lambda_{1}}{r}+\frac{1}{2 \pi \epsilon_{0}} \frac{\lambda_{2}}{r}\\
& =\frac{1}{2 \pi \epsilon_{0} r}\left(\lambda_{1}+\lambda_{2}\right)\\
&=\left(\frac{18 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}}{0.2 \mathrm{m}}\right)(4.8+2.4) 10^{-6} \mathrm{C} / \mathrm{m}\\
&= \boxed{6.5 \times 10^{5} \mathrm{N} / \mathrm{C}}
\end{align}
(b) At y = 0.6 m, both lines apply electric fields in opposite directions, so the net electric field for both kine at y = 0.6 m is calculated by
\begin{aligned}
E&=\frac{1}{2 \pi \epsilon_{0}}\left(\frac{\lambda_{2}}{r_{2}}-\frac{\lambda_{1}}{r_{1}}\right) \\
&=\left(18 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(\frac{2.4 \times 10^{-6} \mathrm{C} / \mathrm{m}}{0.2 \mathrm{m}}-\frac{4.8 \times 10^{-6} \mathrm{C} / \mathrm{m}}{0.6 \mathrm{m}}\right) \\
&=\boxed{7.2 \times 10^{4} \mathrm{N} / \mathrm{C}}
\end{aligned}