The value of Rydberg constant $$(R_H)$$ is $$2.18 \times 10^{-18} \text{ J}$$. The velocity of electron having mass $$9.1 \times 10^{-31} \text{ kg}$$ in Bohr's first orbit of hydrogen atom $$=$$ ______ $$\times 10^5 \text{ ms}^{-1}$$ (nearest integer).

We need to find the velocity of the electron in Bohr's first orbit of the hydrogen atom, given the Rydberg constant and electron mass.

Recall the energy relation in Bohr's model.

In Bohr's model, for the hydrogen atom in the first orbit ($$n = 1$$), the total energy is:

$$ E_1 = -R_H $$

where $$R_H = 2.18 \times 10^{-18}$$ J is the Rydberg constant (energy).

The total energy is the sum of kinetic and potential energies. For a circular orbit under Coulomb's law:

$$ E = -KE = \frac{PE}{2} $$

This gives us: $$KE = -E = R_H$$

Set up the kinetic energy equation.

$$ KE = \frac{1}{2}mv^2 = R_H $$

$$ \frac{1}{2} \times 9.1 \times 10^{-31} \times v^2 = 2.18 \times 10^{-18} $$

Solve for $$v^2$$.

$$ v^2 = \frac{2 \times 2.18 \times 10^{-18}}{9.1 \times 10^{-31}} $$

$$ v^2 = \frac{4.36 \times 10^{-18}}{9.1 \times 10^{-31}} $$

$$ v^2 = \frac{4.36}{9.1} \times 10^{-18+31} = 0.4791 \times 10^{13} = 4.791 \times 10^{12} \text{ m}^2\text{s}^{-2} $$

Calculate $$v$$.

$$ v = \sqrt{4.791 \times 10^{12}} $$

$$ v = \sqrt{4.791} \times 10^6 $$

$$ v \approx 2.189 \times 10^6 \text{ m/s} $$

Express in the required form.

$$ v = 2.189 \times 10^6 = 21.89 \times 10^5 \text{ m/s} $$

Rounding to the nearest integer: $$v \approx 22 \times 10^5$$ m/s.

The answer is $$\boxed{22}$$.