Answer
$8.84\times 10^{5}\;\rm N/C$
Work Step by Step
To understand this problem, we need to sketch it as we see below.
The electrodes here are working as a capacitor (two parallel plates of capacitor).
We know that the electric field inside the capacitor is uniform, so the force exerted by this field on the electron is constant. Hence, the electrons accelerate at a constant rate.
The net force exerted on the ink drop is given by
$$\sum F=qE= (Nm_e+m_d)a_x $$
where $q$ is the charge of a drop, $E$ is the uniform electric field between the plates, $N$ is the number of electrons in one drop, $m_d$ is the mass of the ink drop, $m_e$ is the mass of one electron, and $a_y$ is the acceleration.
Hence,
$$E=\dfrac{Nma_x }{q}$$
where $q=Ne^-$
$$E=\dfrac{ (Nm_e+m_d)a_x }{Ne^-}$$
where the mass of the drop is given by $m_d=\rho V=\frac{4}{3}\rho \pi r^3$
$$E=\dfrac{ (Nm_e+\frac{4}{3}\rho \pi r^3)a_x}{Ne^-}\tag 1$$
Now we need to find $a_x$ to find the magnitude of the electric field.
The drop accelerates in the $x$-direction, while its acceleration in the $y$-direction is zero.
The initial speed in the $x$-direction is zero, so we can find the time of the drop's trip between the two plates by using the constant speed component in the $y$-direction.
$$v_y=\dfrac{L}{t}$$
where $L$ is the length of the plate.
Hence,
$$t=\dfrac{L}{v_y}=\dfrac{6\times 10^{-3}}{20}=\bf 3\times 10^{-4}\;\rm s\tag 2$$
From the geometry of the figure below, we can see that the angle of deflection is given by
$$\tan\theta=\dfrac{3\times 10^{-3}}{2\times 10^{-2}}$$
Hence,
$$\theta={\bf 8.53^\circ\;}\tag 3$$
So at the end of the plates, the drop would be moved a distance of $\Delta x$ which is given by
$$\tan\theta=\dfrac{\Delta x }{0.5L}$$
$0.5L$ since the angle is measured from the center of the plates.
Hence,
$$\Delta x=(0.5L)\tan\theta$$
$$\Delta x=3\tan 8.53^\circ=\bf 0.45\;\rm mm\tag 4$$
Using the kinematic formula of
$$\Delta x=v_{0x}t+\frac{1}{2}a_xt^2=0+\frac{1}{2}a_xt^2$$
So,
$$a_x=\dfrac{2\Delta x}{t^2}=\dfrac{2(0.45\times 10^{-3})}{(3\times 10^{-4})^2}$$
$$a_x=\bf 10,000\;\rm m/s^2$$
Plug all the known into (1),
$$E=\dfrac{ [(8\times 10^5)(9.11\times 10^{-31})+\frac{4}{3}\pi (800) (15\times 10^{-6})^3](10,000)}{((8\times 10^5)(1.6\times 10^{-19})}$$
$$E=\color{red}{\bf 8.84\times 10^{5}}\;\rm N/C$$
