Finding \$L\$ value in simple circuit
Well, first of all: when we've a shirt circuit we know that the resistance (in the switch) is equal to \$0\space\Omega\$ so the resistor \$\text{R}_\text{l}\$ is not important and we can ignore it.

Assuming that the source (that you've put as a current source instead of a voltage supply) is a constant voltage supply of \$100\space\text{V}\$. We can write (according to Faraday's law of induction):

$$-\text{E}\left(t\right)+\text{I}(t)\cdot\text{R}_\text{t}=-\text{I}'(t)\cdot\text{L}\tag1$$

For the initial condition, we have:

$$-\text{E}\left(t\right)+\text{I}_\text{open}(t)\cdot\text{R}_\text{t}+\text{I}(t)_\text{open}\cdot\text{R}_\text{l}=-\text{I}_\text{open}'(t)\cdot\text{L}\space\Longleftrightarrow\space$$
$$\text{I}_\text{open}(t)=\frac{1}{1001}\cdot\left(1-\exp\left(-\frac{100100t}{\text{L}}\right)\right)\tag2$$

So, when \$t=30\cdot10^{-6}\$:

$$\text{I}_\text{open}(30\cdot10^{-6})=\frac{1}{1001}\cdot\left(1-\exp\left(-\frac{100100\cdot30\cdot10^{-6}}{\text{L}}\right)\right)\tag3$$

So, substituting your values we get:

$$\text{I}(30\cdot10^{-6})=20\cdot10^{-3}\space\Longleftrightarrow\space\text{L}\approx0.156224\space\text{H}\tag5$$