The back electromotive force in solenoid is (u) will be maximum. If there is maximum rate of change of current. This happens in $\mathrm{AB}$ part of the graph. Thus maximum back emf will be obtained between $5 \mathrm{~s} < \mathrm{t} < 10$ s. So, the back emf at $\mathrm{t}=3 \mathrm{~s}$ is $\mathrm{e}$.
Also, the rate of change of current at $\mathrm{t}=3$, $\mathrm{s}=$ slope of $\mathrm{OA}$ from $\mathrm{t}=0 \mathrm{~s}$ to $\mathrm{t}=5 \mathrm{~s}=1 / 5 \mathrm{~A} / \mathrm{s}$.
So, back electromotive force
$$
\mathrm{F}=\mathrm{L} \times \frac{1}{5}=\frac{\mathrm{L}}{5}=\mathrm{e}
$$
If $\mathrm{t}=3 \mathrm{~s}, \frac{\mathrm{dI}}{\mathrm{dt}}=1 / 5$ (L is a constant).
Applying $e=L \frac{d I}{d t}$
Similarly, we get back emf $\mathrm{u}_1$ between 5 to $10 \mathrm{sec}$.
For $5 \mathrm{~s} < \mathrm{t} < 10 \mathrm{~s}$
$$
\mathrm{u}_1=-\mathrm{L} \frac{3}{5}=-\frac{3}{5} \mathrm{~L}=-3 \mathrm{e}
$$
So, at $\mathrm{t}=7 \mathrm{~s}, \mathrm{u}_1=-3 \mathrm{e}$
Back emf between 10 to $30 \mathrm{sec}$.
For $10 \mathrm{~s} < \mathrm{t} < 30 \mathrm{~s}$
$$
\mathrm{u}_2=\mathrm{L} \frac{2}{20}=\frac{\mathrm{L}}{10}=\frac{1}{2} \mathrm{e}
$$
For $\mathrm{t}>30 \mathrm{~s}, \mathrm{u}_2=0$
Hence, the back emf at $t=7 \mathrm{~s}, 15 \mathrm{~s}$ and $40 \mathrm{~s}$ are $-3 \mathrm{e}, \mathrm{e} / 2$ and 0 respectively.