Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the electric flux through a closed surface is given by
$$\Phi_e=\oint \vec E\cdot d\vec A$$
So for a closed curve of radius $R$, for a clockwise direction,
$$\Phi_e= EA=E( \pi r^2)$$
For the whole cylinder where $r=R$ where $R$ is the radius of the cylinder.
$$\Phi_e= \pi R^2E \tag 1$$
Plug the known;
$$\Phi_e= \pi (3\times 10^{-3})^2(10^8t^2)$$
$$\boxed{\Phi_e= 900\pi t^2}\tag 2$$
$$\color{blue}{\bf [b]}$$
To find the magnetic field direction, we need to put our right-hand thump finger in the direction of the electric field which is into the page, and curl the other fingers.
It is obvious now that the direction of the magnetic field is clockwise.
Note that we assumed above that the electric flux increases clockwise.
$$\color{blue}{\bf [c]}$$
According to Ampere-Maxwell's law,
$$\oint \vec B\cdot d\vec s=\dfrac{1}{c^2}\dfrac{d\Phi_e}{dt}$$
And for our case, for $r\lt R$,
$$B(2\pi r)=\dfrac{1}{c^2}\dfrac{d\Phi_e}{dt}$$
Solving for $B$, and plug $\Phi_e$ from (1) but replace $R$ by $r$,
$$B =\dfrac{1}{2\pi rc^2}\dfrac{d( \pi r^2E )}{dt}$$
$$B =\dfrac{ \color{red}{\bf\not}\pi r^{ \color{red}{\bf\not}2}}{2 \color{red}{\bf\not}\pi \color{red}{\bf\not}rc^2}\dfrac{dE }{dt}$$
$$B =\dfrac{r}{2 c^2}\dfrac{dE }{dt}$$
where $E=10^8 t^2$
$$B =\dfrac{10^8r}{2 c^2}\dfrac{dt^2 }{dt} =\dfrac{10^8r}{ \color{red}{\bf\not}2 c^2} ( \color{red}{\bf\not}2t)$$
$$\boxed{B =\dfrac{10^8rt}{ c^2} }$$
Plug the known;
$$B =\dfrac{10^8(0.002)(2)}{ (3\times 10^8)^2} $$
$$B =\color{red}{\bf 4.44\times 10^{-12}}\;\rm T $$
$$\color{blue}{\bf [d]}$$
According to Ampere-Maxwell's law,
$$\oint \vec B\cdot d\vec s=\dfrac{1}{c^2}\dfrac{d\Phi_e}{dt}$$
And for our case, for $r\gt R$,
Solving for $B$, and plug $\Phi_e$ from (2) since we deal with $r\gt R$ for the electric flux.
$$B =\dfrac{900 \color{red}{\bf\not}\pi}{2 \color{red}{\bf\not}\pi rc^2}\dfrac{d t^2}{dt}=\dfrac{900 }{ \color{red}{\bf\not}2 rc^2}( \color{red}{\bf\not}2t)$$
$$\boxed{B =\dfrac{900 t}{ rc^2} }$$
Plug the known;
$$B =\dfrac{900 (2)}{ (0.004)(3\times 10^8)^2}$$
$$B =\color{red}{\bf 5.0\times 10^{-12}}\;\rm T $$
