Answer
See the detailed answer below.
Work Step by Step
We know that the electric field inside a capacitor is uniform.
We also know that when the capacitor is charging, the electric field changes during this process.
This change induces a magnetic field.
The author told us the diameter of the plates, so they are circles.
Note that induced magnetic field lines are circles that have the same center as the capacitor plates.
So, according to Ampere-Maxwell law,
$$\oint \vec B\cdot d\vec s=\epsilon_0\mu_0\dfrac{d\Phi_e}{dt}$$
where $\Phi_e=EA$ where $A=\pi r^2$,
$$\oint \vec B\cdot d\vec s=(\pi r^2)\epsilon_0\mu_0\dfrac{d E}{dt}$$
Noting that the magnetic field is tangent to a circle of radius $r$, so $\oint \vec B\cdot d\vec s=2\pi r B$,
$$2\pi r B=(\pi r^2)\epsilon_0\mu_0\dfrac{d E}{dt}\tag 1$$
$$ B= \dfrac{\epsilon_0\mu_0 r}{2}\dfrac{d E}{dt}$$
where $\epsilon_0\mu_0=1/c^2$ where $c$ is the speed of light.
$$ B= \dfrac{ r}{2c^2}\dfrac{d E}{dt}$$
Plug the known;
$$ B= \dfrac{ r}{2(3\times 10^8)^2}(1\times 10^6)$$
$$ \boxed{B= \left( \dfrac{10^{-10}}{ 18}\right) r}$$
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$$\color{blue}{\bf [a]}$$
On the axis means at the center at which $r=0$, plug that into the boxed formula above.
$$ B= \left( \dfrac{10^{-10}}{ 18}\right)(0)=\color{red}{\bf 0}\;\rm T$$
$$\color{blue}{\bf [b]}$$
At $r=0.03$ m, plug that into the boxed formula above.
$$ B= \left( \dfrac{10^{-10}}{ 18}\right)(0.03)=\color{red}{\bf 1.67\times 10^{-13}}\;\rm T$$
$$\color{blue}{\bf [c]}$$
At $r=0.07$ m, which is larger than the radius of the plate of the capacitor, the electric flux is due to the area of the plate only since we assumed that the electric field is uniform inside the capacitor which means it is zero outside it.
Using (1);
$$2\pi r B=(\pi R^2)\epsilon_0\mu_0\dfrac{d E}{dt} $$
$$ B=\dfrac{ R^2}{2r}\epsilon_0\mu_0\dfrac{d E}{dt} $$
$$ B=\dfrac{ R^2}{2rc^2} \dfrac{d E}{dt} $$
Plug the known;
$$ B=\dfrac{ (0.05)^2}{2(0.07)(3\times 10^8)^2} (1\times 10^6)$$
$$ B= \color{red}{\bf 1.98\times 10^{-13}}\;\rm T$$
