Let's first determine the resistance of the coil when connected to a DC supply. The current drawn by the coil in a DC circuit can be used to calculate its resistance using Ohm's law:

$$R = \frac{V}{I}$$

Given the DC supply voltage $$V_{DC} = 20 \, \mathrm{V}$$ and the current $$I_{DC} = 5 \, \mathrm{A}$$, the resistance $$R$$ is:

$$R = \frac{20 \, \mathrm{V}}{5 \, \mathrm{A}} = 4 \, \Omega$$

Next, let's use the information given for the AC supply. When connected to an AC supply, the coil's impedance $$Z$$ can be determined using the given current. The total voltage and current in an AC circuit are related to the impedance by the formula:

$$Z = \frac{V}{I}$$

Given the AC supply voltage $$V_{AC} = 20 \, \mathrm{V}$$ and the current $$I_{AC} = 4 \, \mathrm{A}$$, the impedance $$Z$$ is:

$$Z = \frac{20 \, \mathrm{V}}{4 \, \mathrm{A}} = 5 \, \Omega$$

The impedance $$Z$$ of the coil in an AC circuit is composed of both the resistance $$R$$ and the inductive reactance $$X_L$$, related by:

$$Z = \sqrt{R^2 + X_L^2}$$

We already know that $$R = 4 \, \Omega$$. We can now solve for the inductive reactance $$X_L$$:

$$5 = \sqrt{4^2 + X_L^2}$$

Squaring both sides of the equation:

$$25 = 16 + X_L^2$$

Solving for $$X_L$$:

$$X_L^2 = 25 - 16$$

$$X_L^2 = 9$$

$$X_L = \sqrt{9}$$

$$X_L = 3 \, \Omega$$

The inductive reactance $$X_L$$ is also related to the inductance $$L$$ and the angular frequency $$\omega$$ by the formula:

$$X_L = \omega L$$

where $$\omega = 2 \pi f$$. Given the frequency $$f = 50 \, \mathrm{Hz}$$ and using $$\pi = 3$$, we find:

$$\omega = 2 \times 3 \times 50$$

$$\omega = 300 \, \mathrm{rad/s}$$

Now we can solve for the inductance $$L$$:

$$X_L = 300 L$$

$$3 = 300 L$$

$$L = \frac{3}{300}$$

$$L = 0.01 \, \mathrm{H}$$

Since $$1 \, \mathrm{H} = 1000 \, \mathrm{mH}$$, the self inductance of the coil is:

$$L = 0.01 \, \mathrm{H} \times 1000 \, \mathrm{mH/H} = 10 \, \mathrm{mH}$$

Therefore, the self inductance of the coil is $$10 \, \mathrm{mH}$$.