To solve this problem, note that when a charged particle with mass $$m$$ and charge $$q$$ is fired perpendicular to a magnetic field of strength $$B$$, it undergoes uniform circular motion. The period (time to complete one full circle) is given by:

$$ T = \frac{2\pi m}{qB} $$

Here's how to calculate it step by step:

Convert the given quantities to SI units:

Charge: $$q = 1.6\,\mu\text{C} = 1.6 \times 10^{-6}\,\text{C}$$

Mass: $$m = 16\,\mu\text{g} = 16 \times 10^{-9}\,\text{kg}$$

Magnetic field: $$B = 6.28\,\text{T}$$

Write down the period formula:

$$ T = \frac{2\pi m}{qB} $$

Substitute the values:

$$ T = \frac{2\pi \times (16 \times 10^{-9}\,\text{kg})}{(1.6 \times 10^{-6}\,\text{C})(6.28\,\text{T})} $$

Notice that $$2\pi \approx 6.28$$, which cancels with the given magnetic field value, simplifying the expression:

$$ T = \frac{6.28 \times 16 \times 10^{-9}}{1.6 \times 10^{-6} \times 6.28} = \frac{16 \times 10^{-9}}{1.6 \times 10^{-6}} $$

Simplify the fraction:

$$ T = \frac{16}{1.6} \times \frac{10^{-9}}{10^{-6}} = 10 \times 10^{-3} = 0.01\,\text{s} $$

Thus, the time required for the particle to return to its original location is:

$$ \boxed{0.01\,\text{s}} $$