Let us consider an infinitesimal element of length dx at a distance x from the point P.

The charge on this element is dq = λdx,
where, $\mathrm{\lambda } = \frac{\mathrm{Q}}{\mathrm{L}}$ is the linear charge density.

The magnitude of the electric field at P due to this element is
         $$\begin{gathered} \mathrm{dE} = \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{dq}}{{\mathrm{x}}^2} \\ = \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(\mathrm{\lambda dx}\right)}{{\mathrm{x}}^2} \end{gathered}$$

and its direction is to the right since X is positive.
The total electric field strength E is given by
$\mathrm{E} = \frac{1}{4{\mathrm{\pi \epsilon }}_0}\mathrm{\lambda } {\int }_{\mathrm{a}}^{\mathrm{a}+\mathrm{L}}\frac{\mathrm{dx}}{{\mathrm{x}}^2}$

$$\begin{gathered} = \frac{\mathrm{\lambda }}{4{\mathrm{\pi \epsilon }}_0}{\left[-\frac{1}{\mathrm{x}}\right]}_{\mathrm{a}}^{\mathrm{a}+\mathrm{L}} \\ = \frac{\mathrm{\lambda }}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{a}+\mathrm{L}}\right] \\ = \frac{\mathrm{Q}}{\left(4{\mathrm{\pi \epsilon }}_0\right) \mathrm{a} (\mathrm{a}+\mathrm{L})} (\therefore \mathrm{Q} = \mathrm{\lambda L}) \end{gathered}$$

Given, 

Distance between the charges, r = $3.2\times {10}^{-3} m$
 $$\begin{gathered} \mathrm{Iinitial} \mathrm{acceleration} \mathrm{of} \mathrm{first} \mathrm{particle}, {\mathrm{a}}_1 = 7.0 \mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Initial} \mathrm{acceleration} \mathrm{of} \mathrm{second} \mathrm{particle}, {\mathrm{a}}_2 = 9.0 \mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Mass} \mathrm{of} \mathrm{first} \mathrm{particle}, {\mathrm{m}}_1 = 6.3 \times {10}^{-7} \mathrm{kg}, \\ \mathrm{Mas} \mathrm{of} \mathrm{second} \mathrm{particle}, {\mathrm{m}}_2 = ? \end{gathered}$$

a) Since,   ${\mathrm{F}}_1 = {\mathrm{F}}_2$

$\therefore$        ${\mathrm{m}}_1{\mathrm{a}}_1 = {\mathrm{m}}_2{\mathrm{a}}_2$

   $$\begin{gathered} \mathrm{mass} \mathrm{of} \mathrm{second} \mathrm{particle}-{\mathrm{m}}_2 = \frac{{\mathrm{m}}_1{\mathrm{a}}_1}{{\mathrm{a}}_2} \\ = \frac{6.3\times {10}^{-7}\times 7.0}{9.0} \\ = 4.9 \times {10}^{-7} \mathrm{kg} \end{gathered}$$ 

b)As,
                $$\begin{gathered} {\mathrm{F}}_1 = {\mathrm{F}}_2 \\ = \frac{{\mathrm{q}}_1 {\mathrm{q}}_2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2} = {\mathrm{m}}_1{\mathrm{a}}_1 \\ = 6.3 \times {10}^{-7} \times 7.0 \\ = 44.1 \times {10}^{-7} \end{gathered}$$ 

$\therefore$             $\frac{9\times {10}^9 {\mathrm{q}}^2}{{\left(3.2 \times {10}^{-3}\right)}^2} = 44.1 \times {10}^{-7}$ 

$\therefore$                $\mathrm{q} = 7.1 \times {10}^{-11}\mathrm{C}.$