Answer
See the detailed answer below.
Work Step by Step
We know that the magnetic field of a solenoid is given by
$$B_{\rm solenoid}=\dfrac{\mu_0 NI}{l}\tag 1$$
And we know that the strength of the induced electric field inside a solenoid is given by
$$E_{\rm solenoid}=\dfrac{r}{2}\left| \dfrac{dB}{dt} \right|$$
Plug $B$ from (1) and note that the only changing variable in this case is the current $I$,
$$E_{\rm solenoid}=\dfrac{\mu_0 Nr}{2l}\left| \dfrac{dI}{dt} \right| $$
Plug the known;
$$E_{\rm solenoid}=\dfrac{(4\pi \times 10^{-7})(400)(0.01)}{2(0.2)}\left| \dfrac{dI}{dt} \right| $$
$$E_{\rm solenoid}=(1.257\times 10^{-5})\left| \dfrac{dI}{dt} \right|\tag 2$$
We have 3 stages here, as we see from the given graph:
$\bullet$ At stage 1, from 0 s to 0.1 s: $dI/dt=\rm Slope_1=50\; A/s$.
Plug into (2),
$$E_{\rm solenoid,1}=(1.257\times 10^{-5})\left| 50 \right| $$
$$E_{\rm solenoid,1}=\color{red}{\bf 6.28\times 10^{-4}}\;\rm V/m$$
$\bullet$ $\bullet$ At stage 2, from 0.1 s to 0.2 s: $dI/dt=\rm Slope_1=0\; A/s$.
Plug into (2),
$$E_{\rm solenoid,2}=(1.257\times 10^{-5})\left| 0\right| $$
$$E_{\rm solenoid,2}=\color{red}{\bf0 }\;\rm V/m$$
$\bullet$ $\bullet$ $\bullet$ At stage 3, from 0.2 s to 0.4 s: $dI/dt=\rm Slope_1=25\; A/s$.
Plug into (2),
$$E_{\rm solenoid,3}=(1.257\times 10^{-5})\left| 25 \right| $$
$$E_{\rm solenoid,3}=\color{red}{\bf 3.14\times 10^{-4}}\;\rm V/m$$
Now we can easily draw the need graph of $E$ versus $t$, as shown below.
