To find the self-inductance of the coil, we can use the formula for induced electromotive force (emf), which is given by Faraday's law of electromagnetic induction as it applies to self-induction:

$$ \text{emf} = - L \frac{\Delta I}{\Delta t} $$

Where:

• $ \text{emf} $ = induced voltage in volts (V)
• $ L $ = self-inductance of the coil in henries (H)
• $ \Delta I $ = change in current in amperes (A)
• $ \Delta t $ = time interval in seconds (s) over which the current change occurs

Here, the problem gives us the following data:

• $ \text{emf} = 0.1 \, \text{V} $
• $ \Delta I = 2 \, \text{A} - (-2 \, \text{A}) = 4 \, \text{A} $
• $ \Delta t = 0.2 \, \text{s} $

Substituting the given values into the formula, we get:

$$ 0.1 = -L \frac{4}{0.2} $$

Solving for $ L $, the equation becomes:

$$ 0.1 = -L \cdot 20 $$

Therefore:

$$ L = - \frac{0.1}{20} $$

Calculating the value of $ L $:

$$ L = -0.005 \, \text{H} $$

Or, expressing $ L $ in millihenries (mH):

$$ L = -5 \, \text{mH} $$

However, considering the absolute value (since inductance is a magnitude and cannot be negative in this context):

$$ L = 5 \, \text{mH} $$

Thus, the self-inductance of the coil is 5 mH, which corresponds to Option D.