Answer
$6.56\times10^{12}\;\rm m/s^2$
Work Step by Step
The two plates will accelerate the ions in the $y$-direction so that the horizontal velocity component remains constant.
$$v_x=5\times10^6\;\rm m/s\tag 1$$
We chose the left side of the plates to be out origin $(0,0)$
The intil vertical velocity component is zero since it starts to accelerate up from the left side of the plates.
$$v_{y1}=v_{0y}+a_yt_1=0+a_yt_1$$
$$v_{y1}=a_yt_1\tag 2$$
whereas $t_1$ is the interval time of acceleration.
Also,
$$v_{y1}^2=v_{iy}^2+2a_y\Delta x_1=0+2a_y\Delta x_1$$
$$v_{y1}^2= 2a_y\Delta x_1 \tag 3$$
whereas $\Delta x_1=5\;\rm cm$ which is the acceleration distance.
To find the acceleration, we need to find $t_1$ and $v_{y1}$.
We can find $t_1$ from
$$v_x=\dfrac{\Delta x_1}{t_1}$$
Thus,
$$t_1 =\dfrac{\Delta x_1}{v_x}=\dfrac{5\times10^{-2}}{5\times10^6}= \color{blue}{1.0\times10^{-8}}\;\rm s$$
We also can find the time interval of the second stage of motion, from the right side of the plates to the tissue,
$$t_2=\dfrac{\Delta x_2}{v_x}=\dfrac{1.5}{5\times10^6}=\color{blue}{3\times10^{-7}}\;\rm s$$
Now we need to use the kinematic formula of
$$y_1=\overbrace{y_0}^{0}+\overbrace{v_{0y}t_1}^{0}+\frac{1}{2}a_yt_1^2$$
to find the vertical distance traveled when the ion was still between the plates.
Thus,
$$y_1=\frac{1}{2}a_yt_1^2\tag 4$$
and the vertical distance traveled outside the paltes
$$y_2 =y_1+v_{y1}t_2$$
Plugging from (2) and (4)
$$2\times10^{-2}=\frac{1}{2}a_yt_1^2+a_yt_1t_2$$
$$2\times10^{-2}=a_y\left[\frac{1}{2} t_1^2+ t_1t_2\right]$$
$$a_y=\dfrac{2\times10^{-2}}{\frac{1}{2} t_1^2+ t_1t_2}$$
$$a_y=\dfrac{2\times10^{-2}}{\frac{1}{2} \left[1.0\times10^{-8}\right]^2+\left[1.0\times10^{-8}\cdot 3\times10^{-7}\right]} $$
$$\boxed{a_y=\color{red}{\bf 6.56\times10^{12}}\;\rm m/s^2}$$
