To convert a galvanometer into an ammeter, we need to add a shunt resistance in parallel with the galvanometer's coil. The purpose of the shunt resistance is to bypass the majority of the current while allowing only a small fraction of it to pass through the galvanometer, thereby preventing it from being damaged by high currents.

Given parameters:

• Resistance of the galvanometer's coil, $$R_g = 200 \Omega$$


• Full-scale deflection current of the galvanometer, $$I_g = 20 \mu \mathrm{A} = 20 \times 10^{-6} \mathrm{A}$$


• Desired ammeter range, $$I = 20 \mathrm{mA} = 20 \times 10^{-3} \mathrm{A}$$

The shunt resistance, $$R_s$$, can be calculated using the formula:

$$ \frac{R_s}{R_g + R_s} = \frac{I_g}{I} $$

Solving for $$R_s$$:

$$ R_s = R_g \left(\frac{I_g}{I - I_g}\right) $$

Substituting the given values:

$$ R_s = 200 \left(\frac{20 \times 10^{-6}}{20 \times 10^{-3} - 20 \times 10^{-6}}\right) $$

Simplifying the expression:

$$ R_s = 200 \left(\frac{20 \times 10^{-6}}{19.98 \times 10^{-3}}\right) $$

$$ R_s \approx 200 \left(\frac{20 \times 10^{-6}}{20 \times 10^{-3}}\right) = 200 \left(\frac{1}{1000}\right) = 0.20 \Omega $$

Therefore, the value of the resistance to be added to use the galvanometer as an ammeter of range $$(0-20) \mathrm{mA}$$ is $$0.20 \Omega$$. Thus, the correct answer is:

Option C: $$0.20 \Omega$$