Answer
See the detailed answer below.
Work Step by Step
After moving the switch from a to b, the battery is no longer connected and the current moves through the $RL$ circuit.
Before moving the switch, the current is constant and the ideal inductor is assumed to work as a wire with no resistance.
Thus the instantaneous current after moving the switch is
$$I_0=\dfrac{V_B}{R}=\dfrac{9}{100}=\bf 0.09\;\rm A\tag 1$$
The current starts to decay and this decay is given
$$I=I_0e^{-tR/L}$$
where $I=V_R/R$
$$\dfrac{V_R}{R}=\dfrac{(V_R)_0}{R}e^{-tR/L}$$
Hence,
$$ V_R = (V_R)_0 e^{-tR/L}$$
$$\dfrac{ V_R}{ (V_R)_0}= e^{-tR/L}$$
So,
$$ \ln\left[\dfrac{ V_R}{ (V_R)_0}\right]= \dfrac{ -R}{L}t $$
where $ (V_R)_0=I_0R=(0.09)(100)=\bf 9\;\rm V$
$$\boxed{\ln\left[\frac{1 }{ 9}V_R\right]= \dfrac{ -R}{L}t}$$
Now we have a straight-line equation $y=mx+b$ where:
- $y=\ln\left[\dfrac{ V_R}{ (V_R)_0}\right]$
- $m={\rm slope}=\dfrac{-R}{L}$
- $x=t$
- $b=0$
So if we draw this line and find its slope, we can find $L$ where
$${\rm slope}=\dfrac{-R}{L}$$
Thus,
$$L=\dfrac{-R}{{\rm slope}}\tag 3$$
Now we need to plug the given data into the boxed formula above and then plug the resulting dots in the graph, as we see in the graph below.
And from it, it is obvious that the slope is $-0.03\times 10^6$ plug that into (3),
$$L=\dfrac{-R}{-0.03\times 10^6}=\dfrac{-100}{-0.03\times 10^6} =\color{red}{\bf 3.3}\;\rm mH$$
