Answer
See the graph below,
Work Step by Step
The displacement current through the capacitor is given by
$$I_{\rm disp }=\dfrac{dQ}{dt}$$
Recalling that $V_C=Q/C\Rightarrow Q=V_CC$,
$$I_{\rm disp }=C\dfrac{dV_C}{dt}$$
where $dV_C/dt$ is the slope of the given graph where we have three stages there.
$$I_{\rm disp }=C \cdot{\rm slope}$$
Plug the known;
$$I_{\rm disp }=(0.10\times 10^{-6}) \cdot{\rm slope}\tag 1$$
$*$ For the first stage, the slope is 100,
Plug into (1),
$$I_{\rm disp }=(0.10\times 10^{-6}) \cdot (100)=\bf 10\;\rm \mu A$$
$**$ For the second stage, the slope is 0,
Plug into (1),
$$I_{\rm disp }=(0.10\times 10^{-6}) \cdot (0)=\bf 0\;\rm \mu A$$
$***$ For the third stage, the slope is -50,
Plug into (1),
$$I_{\rm disp }=(0.10\times 10^{-6}) \cdot (-50)=\bf -5.0\;\rm \mu A$$
Now we can draw the displacement current as a function of time.
See the figure below
