To convert a galvanometer into an ammeter to measure larger currents, a low resistance known as the shunt resistance ($$R_{\text{sh}}$$) is connected in parallel with the galvanometer. The value of this shunt resistance can be calculated using the principles of parallel circuits and the desired maximum current the ammeter should read.

The original configuration of the galvanometer allows it to measure up to $$10\,\text{V}$$, and it has a resistance of $$100\,\Omega$$. When connected in series with a $$400\,\Omega$$ resistor, the total resistance in the circuit is $$100\,\Omega + 400\,\Omega = 500\,\Omega$$. Given this configuration measures up to $$10\,\text{V}$$, we can calculate the maximum current it is designed to measure using Ohm's law:

$I = \frac{V}{R} = \frac{10\,\text{V}}{500\,\Omega} = 0.02\,\text{A}$

Now, to recalibrate the device to measure up to $$10\,\text{A}$$, we require the calculation of the shunt resistor $$R_{\text{sh}}$$ that needs to be connected in parallel with the galvanometer. The total current $$I$$ will now be $$10\,\text{A}$$, and the part of this current flowing through the galvanometer ($$I_g$$) remains $$0.02\,\text{A}$$ (as before, to ensure we do not exceed the device's original maximum measuring capability), leaving the rest to flow through the shunt. Thus, $$I - I_g$$ flows through the shunt.

Since the voltage drop across both the shunt and the galvanometer must be the same for parallel components, we use Ohm’s Law $$V = IR$$ for both and set up an equation to calculate $$R_{\text{sh}}$$:

$I_g R_g = (I - I_g) R_{\text{sh}}$

Substituting known values ($$R_g = 100\,\Omega$$, $$I = 10\,\text{A}$$, and $$I_g = 0.02\,\text{A}$$):

$0.02\,\text{A} \times 100\,\Omega = (10\,\text{A} - 0.02\,\text{A}) R_{\text{sh}}$

This simplifies to:

$2\,\text{V} = 9.98\,\text{A} \times R_{\text{sh}}$

Solving for $$R_{\text{sh}}$$ gives:

$R_{\text{sh}} = \frac{2\,\text{V}}{9.98\,\text{A}} \approx 0.2004\,\Omega$

Expressing this in terms of $$\times 10^{-2} \Omega$$ gives $$R_{\text{sh}} \approx 20.04 \times 10^{-2} \Omega$$. Therefore, the value of $$x$$ is approximately $$20.04$$, and rounding it according to the provided options leads to the closest value:

Option B: 20