Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the displacement current is given by
$$I_{\rm disp}=\epsilon_0\dfrac{d\Phi_e}{dt}\tag 1$$
Recalling that the current density is given by
$$J=\dfrac{I}{A}=\sigma E$$
So,
$$I=A\sigma E$$
And for a current rate $dI/dt$,
$$\dfrac{dI}{dt}=\dfrac{d}{dt}(A\sigma E)$$
where $\sigma$ is a constant,
$$\dfrac{dI}{dt}=\sigma\dfrac{d}{dt}(A E)$$
Recalling that $\Phi_e=AE$, hence
$$\dfrac{dI}{dt}=\sigma\dfrac{d\Phi_e}{dt} $$
Plug from (1),
$$\dfrac{dI}{dt}=\sigma\dfrac{I_{\rm disp}}{\epsilon_0} $$
Thus,
$$\boxed{I_{\rm disp}=\dfrac{\epsilon_0}{\sigma}\dfrac{dI}{dt}}$$
$$\color{blue}{\bf [b]}$$
Plug the known into the boxed formula above.
Recall that $\sigma$ here is for copper.
$$ I_{\rm disp}=\dfrac{(8.85\times 10^{-12})}{(6\times 10^7)}(1\times 10^6) $$
$$ I_{\rm disp}=\color{red}{\bf 1.5\times 10^{-13}}\;\rm A$$
