Answer
See the detailed answer below.
Work Step by Step
We have here a current carrying wire when the switch is closed.
The magnetic field exerted by the wire at the given point is given by
$$B=\dfrac{\mu_0I}{2\pi d}\tag 1$$
When the switch is closed, we have a simple $ RC$ circuit where the capacitor discharges.
So the current is given by
$$I=I_{\rm max}e^{-t/RC}$$
where $I_{\rm max}=\dfrac{(\Delta V_C)_0}{R}$,
$$I=\dfrac{\Delta V_C}{R}e^{-t/RC}$$
Plug into (1),
$$B=\dfrac{\mu_0 }{2\pi d}\dfrac{\Delta V_C}{R}e^{-t/RC}$$
Plug the known;
$$B=\dfrac{(4\pi \times 10^{-7})}{2\pi (0.01)}\dfrac{50}{5}e^{-t/(5)(2\times 10^{-6})}$$
$$\boxed{B=0.0002e^{-t/(5)(2\times 10^{-6})}}$$
Now we can draw the graph, as seen below.
We can see that the maximum field strength occurs at the $y$-intercept point at which
$$B_{\rm max}=\color{red}{\bf 2\times 10^{-4}}\;\rm T$$
