Answer
$F = 0.081 \mathrm{N}$
Work Step by Step
We will calacute the force due to the applied electric field on the charge where the charge $q$ is equal
$$q=\lambda X = (5.2 \times 10^{-6} \mathrm{C} / \mathrm{m})(0.0500 \mathrm{m})=2.6 \times 10^{-7} \mathrm{C}$$
\begin{align*}
F&=\frac{1}{2 \pi \epsilon_{o}} \frac{\lambda}{r} (q) \\
&=\frac{5.2 \times 10^{-6} \mathrm{C} / \mathrm{m}}{2 \pi\left(8.854 \times 10^{-12} \mathrm{C}^{2} /\mathrm{N} \cdot \mathrm{m}^{2}\right)(0.3 \mathrm{m})} (2.6 \times 10^{-7} \mathrm{C})\\
&= \boxed{0.081 \mathrm{N}}
\end{align*}