To determine the difference between the total mass of all the nucleons and the nuclear mass of the given nucleus, we need to use the concept of mass-energy equivalence provided by Einstein's famous equation:

$$ E= \Delta m \cdot c^2 $$

where:

• E is the binding energy.
• $$ \Delta m $$ is the mass defect (difference in mass).
• $$ c $$ is the speed of light in vacuum, which is approximately $$3 \times 10^8 \, \text{m/s}$$.

Given:

• Binding energy, $$ E = 18 \times 10^8 \, \text{J} $$.

We need to find $$ \Delta m $$, so rearranging the equation:

$$ \Delta m = \frac{E}{c^2} $$

Substitute the given values:

$$ \Delta m = \frac{18 \times 10^8}{(3 \times 10^8)^2} $$

Calculate the value:

$$ \Delta m = \frac{18 \times 10^8}{9 \times 10^{16}} $$

$$ \Delta m = \frac{18}{9} \times 10^{-8} $$

$$ \Delta m = 2 \times 10^{-8} \, \text{kg} $$

To convert the mass from kilograms to micrograms ($$ \mu g $$), we use the conversion factor $$ 1 \, \text{kg} = 10^9 \, \mu g $$:

$$ \Delta m = 2 \times 10^{-8} \, \text{kg} \times 10^9 \, \frac{\mu g}{\text{kg}} $$

$$ \Delta m = 2 \times 10^{1} \, \mu g $$

$$ \Delta m = 20 \, \mu g $$

Therefore, the difference between the total mass of all the nucleons and the nuclear mass of the given nucleus is 20 $$ \mu g $$.

Option A (20 $$ \mu g $$) is the correct answer.