Answer
See the detailed answer below.
Work Step by Step
Let's assume that the given area is a cross-sectional area of a solenoid (since it looks like it).
We know that the strength of the induced electric field inside a solenoid is given by
$$E =\dfrac{r}{2}\left| \dfrac{dB}{dt} \right|\tag 1$$
And we know that the electric force exerted on an electron is given by
$$F=eE =ma$$
Hence,
$$E=\dfrac{ma}{e}$$
Plug into (1),
$$\dfrac{ma}{e}=\dfrac{r}{2}\left| \dfrac{dB}{dt} \right| $$
Solving for $a$,
$$a=\dfrac{er}{2m}\left| \dfrac{dB}{dt} \right| \tag 2$$
At point a, $r=0.01$ m plug that, and the rest of the known, into (2)
$$a_a=\dfrac{(1.6\times 10^{-19})(0.01)}{2(1.67\times 10^{-27})}\left| -0.10 \right| $$
$$a_a=\color{red}{\bf 4.8\times 10^4}\;\rm m/s^2\tag{Upward} $$
The direction of the proton is in the same direction as the induced current, and the induced current at point (a) must be clockwise to oppose the decrease of the flux. Hence, the charge (proton) had to move up.
Let's assume that each proton at these 4 positions will move in an imaginary loop that its center is at point b.
At point b, $r=0$ m plug that, and the rest of the known, into (2)
$$a_a=\dfrac{(1.6\times 10^{-19})(0)}{2(1.67\times 10^{-27})}\left| -0.10 \right| $$
$$a_a=\color{red}{\bf 0}\;\rm m/s^2 $$
The proton remains stationary.
At point c, $r=0.01$ m plug that, and the rest of the known, into (2)
$$a_a=\dfrac{(1.6\times 10^{-19})(0.01)}{2(1.67\times 10^{-27})}\left| -0.10 \right| $$
$$a_a=\color{red}{\bf 4.8\times 10^4}\;\rm m/s^2\tag{Downward} $$
The direction of the proton is in the same direction as the induced current, and the induced current at point (c) must be clockwise to oppose the decrease of the flux. Hence, the charge (proton) had to move down.
At point d, $r=0.02$ m plug that, and the rest of the known, into (2)
$$a_a=\dfrac{(1.6\times 10^{-19})(0.02)}{2(1.67\times 10^{-27})}\left| -0.10 \right| $$
$$a_a=\color{red}{\bf 9.6\times 10^4}\;\rm m/s^2\tag{Downward} $$
The direction of the proton is in the same direction as the induced current, and the induced current at point (c) must be clockwise to oppose the decrease of the flux. Hence, the charge (proton) had to move down.
