1. Initial steady current in the coil

A $12\,\text{V}$ battery is connected to a coil of resistance $6\,\Omega$. At steady state, the current is

$I = \frac{V}{R} = \frac{12}{6} = 2\,\text{A}.$

2. Current change when switch is opened

When the switch is opened, the current falls from $2\,\text{A}$ to $0$ in

$\Delta t = 1\,\text{ms} = 10^{-3}\,\text{s}.$

So,

$\left|\frac{\Delta I}{\Delta t}\right| = \frac{2}{10^{-3}} = 2000\,\text{A/s}.$

3. Use formula for induced emf

Magnitude of induced emf across an inductor is

$|\mathcal{E}| = L\left|\frac{dI}{dt}\right|.$

Given induced emf is $20\,\text{V}$, so

$20 = L \cdot 2000.$

Thus,

$L = \frac{20}{2000} = 0.01\,\text{H} = 10\,\text{mH}.$

4. Match with options

$10\,\text{mH}$ corresponds to Option C.

5. Comparison with stored answer

Stored correct answer is C, which matches our derived result.