A thin infinite sheet charge and an infinite line charge of...

Let's consider the following diagram:

The formula to calculate the electric field due to an infinitely long charged sheet is given by

${E}_{s} = \dfrac{\sigma}{2{ε}_{0}} . . . \left(1\right)$

The formula to calculate the electric field due to an infinitely long line charge is given by

${E}_{l} = \dfrac{\lambda}{2\pi{ε}_{0}r} . . . \left(2\right)$

Using equations (1) and (2), the electric field at $P$ can be calculated as follows:

$\begin{cases}{E}_{P} & = & \left|\dfrac{\sigma}{2{ε}_{0}} - \dfrac{1}{4\pi{ε}_{0}}\dfrac{2\lambda}{\dfrac{3}{\pi}}\right| \\ & = & \left|\dfrac{\sigma}{2{ε}_{0}} - \dfrac{\lambda}{6{ε}_{0}}\right| \\ & = & \dfrac{\sigma}{6{ε}_{0}} . . . \left(3\right)\end{cases}$

Similarly, the electric field at $Q$ is given by

$\begin{cases}{E}_{Q} & = & \left|\dfrac{\sigma}{2{ε}_{0}} - \dfrac{1}{4\pi{ε}_{0}}\dfrac{2\lambda}{\dfrac{4}{\pi}}\right| \\ & = & \left|\dfrac{\sigma}{2{ε}_{0}} - \dfrac{\lambda}{8{ε}_{0}}\right| \\ & = & \dfrac{\sigma}{4{ε}_{0}} . . . \left(4\right)\end{cases}$

Divide equation (3) by equation (4) to obtain the required ratio.

$\begin{cases}\dfrac{{E}_{P}}{{E}_{Q}} & = & \dfrac{\dfrac{\sigma}{6{ε}_{0}}}{\dfrac{\sigma}{4{ε}_{0}}} \\ & = & \dfrac{4}{6} \\ & & \end{cases}$

Hence, $a = 6$.