The two capacitors depicted in the figure consist of square plates...

The two capacitors are connected in series. Let their capacitances be $C_1 = \dfrac{\varepsilon_0 a^2}{d_1}$ and $C_2 = \dfrac{\varepsilon_0 a^2}{d_2}$.

The potential difference across the upper capacitor is given by $V_1 = V \dfrac{C_2}{C_1 + C_2}$.

Substituting the expressions for capacitance, we get:
$V_1 = V \dfrac{\dfrac{\varepsilon_0 a^2}{d_2}}{\dfrac{\varepsilon_0 a^2}{d_1} + \dfrac{\varepsilon_0 a^2}{d_2}} = V \dfrac{d_1}{d_1 + d_2}$

The electric field between the plates of the upper capacitor is:
$E_1 = \dfrac{V_1}{d_1} = \dfrac{V}{d_1 + d_2}$

The acceleration of the electron in the vertical direction is:
$a_y = \dfrac{e E_1}{m} = \dfrac{e V}{m (d_1 + d_2)}$

The time taken by the electron to cross the plates of length $a$ with horizontal speed $u$ is $t = \dfrac{a}{u}$.

For the electron to just avoid colliding with the plates, its maximum vertical deflection should be $y = \dfrac{d_1}{2}$.

Using the equation of motion $y = \dfrac{1}{2} a_y t^2$, we have:
$\dfrac{d_1}{2} = \dfrac{1}{2} \left( \dfrac{e V}{m (d_1 + d_2)} \right) \left( \dfrac{a}{u} \right)^2$

$d_1 = \dfrac{e V a^2}{m u^2 (d_1 + d_2)}$

$u^2 = \dfrac{e V a^2}{m d_1 (d_1 + d_2)}$

$u = \sqrt{\dfrac{e V a^2}{m d_1 (d_1 + d_2)}}$