The energy of an electron in a hydrogen atom is given by the following formula:

$$E_n = \frac{-13.6\, \mathrm{eV}}{n^2}$$

where n is the principal quantum number (Bohr orbit number). To find the energy in the third Bohr orbit (n = 3), we can use the formula:

$$E_3 = \frac{-13.6\, \mathrm{eV}}{3^2} = \frac{-13.6\, \mathrm{eV}}{9}$$

Now, we are given the energy of the electron in the first Bohr orbit (n = 1):

$$E_1 = -2.18 \times 10^{-18}\, \mathrm{J}$$

First, let's convert this energy to electron volts (eV):

$$E_1 = -2.18 \times 10^{-18}\, \mathrm{J} \times \frac{1\, \mathrm{eV}}{1.6 \times 10^{-19}\, \mathrm{J}} \approx -13.6\, \mathrm{eV}$$

Now we can find the ratio of $$E_3$$ to $$E_1$$:

$$\text{Ratio} = \frac{E_3}{E_1} = \frac{-\frac{13.6\, \mathrm{eV}}{9}}{-13.6\, \mathrm{eV}} = \frac{1}{9}$$

Thus, the energy of the electron in the third Bohr orbit is $$\frac{1}{9}$$ th of its energy in the first Bohr orbit.