1. Use Bohr model energy relation

For hydrogen atom, the total energy in the first orbit is $E_1=-2.18\times 10^{-18}\,\text{J}$

The given value called Rydberg constant here is actually the magnitude of ground-state energy.

In Bohr model, $K = -E$ where $K$ is kinetic energy.

So, $K = 2.18\times 10^{-18}\,\text{J}$

2. Write kinetic energy in terms of velocity

$K=\frac{1}{2}mv^2$

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

3. Solve for $v^2$

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

$v^2=\frac{4.36}{9.1}\times 10^{13}$

$v^2\approx 0.4791\times 10^{13}$

$v^2\approx 4.791\times 10^{12}$

4. Take square root

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

$v\approx 2.19\times 10^6\,\text{m s}^{-1}$

5. Express in the required form

We need $v = (\text{number})\times 10^5\,\text{m s}^{-1}$

So, $2.19\times 10^6 = 21.9\times 10^5$

Nearest integer $=22$.

Final Answer

$\boxed{22}$