Answer
See the detailed answer below.
Work Step by Step
We know that the induced current is given by
$$I=\dfrac{\varepsilon}{R}\tag 1$$
where the induced emf $\varepsilon$ is given by
$$\varepsilon=\left|\dfrac{d\Phi}{dt}\right|$$
where $\phi =AB\cos\theta$ and here $A$ is constant and $\theta= 0^\circ$, so that
$$\varepsilon=A\left|\dfrac{dB}{dt}\right|$$
$$\varepsilon=\pi r^2\left|\dfrac{ dB}{dt}\right|\tag 2$$
$$\color{blue}{\bf [a]}$$
Plug the known, from the given graph, into (2) to find the induced emf.
$$\varepsilon=\pi (0.05)^2\left|+0.50\right|=\color{red}{\bf 3.93}\;\rm mV$$
Plug the known into (1) to find the induced current,
$$I=\dfrac{3.93\times 10^{-3}}{0.2} $$
$$I=\color{red}{\bf 20}\;\rm mA\tag{CCW}$$
The external magnetic field increases into the page, so the induced current should create an induced field that opposes this increase which means out of the page.
Thus, according to the right-hand rule, the direction of the induced current is counterclockwise.
$$\color{blue}{\bf [b]}$$
Plug the known, from the given graph, into (2) to find the induced emf.
$$\varepsilon=\pi (0.05)^2\left|-0.50\right|=\color{red}{\bf 3.93}\;\rm mV$$
Plug the known into (1) to find the induced current,
$$I=\dfrac{3.93\times 10^{-3}}{0.2} $$
$$I=\color{red}{\bf 20}\;\rm mA\tag{CW}$$
The external magnetic field decreases out of the page, so the induced current should create an induced field that opposes this increase which means into the page.
Thus, according to the right-hand rule, the direction of the induced current is clockwise.
$$\color{blue}{\bf [c]}$$
The induced emf $\varepsilon$ is given by
$$\varepsilon=\left|\dfrac{d\Phi}{dt}\right| $$
where $\phi =AB\cos\theta$ and here $A$ is constant and $\theta=90^\circ$, so that
$$\varepsilon=A\left|\dfrac{dB}{dt}\right| \cos90^\circ$$
$$\varepsilon=\color{red}{\bf 0}\;\rm V$$
And hence there is no induced current.
$$I=\color{red}{\bf 0}\;\rm A $$
