The radius of the second Bohr orbit for hydrogen atom is
(Planck's constant, (h) = $$6.6262 \times 10^{-34}$$ Js; mass of electron = $$9.1091 \times 10^{-31}$$ kg; charge of electron = $$1.60210 \times 10^{-19}$$ C; permittivity of vacuum, ($$\epsilon_{0}$$) = $$8.854185 \times 10^{-12}$$ kg$$^{-1}$$ m$$^{-3}$$ A$$^{2}$$)

We begin with Bohr’s postulate for the stationary orbits of the hydrogen atom. The radius of the $$n^{\text{th}}$$ orbit is given by the expression

$$ r_n \;=\; \frac{\varepsilon_0 h^2 n^2}{\pi m e^2}, $$

where

$$\varepsilon_0 = 8.854185 \times 10^{-12}\;\text{kg}^{-1}\,\text{m}^{-3}\,\text{A}^2,$$
$$h = 6.6262 \times 10^{-34}\;\text{J\,s},$$
$$m = 9.1091 \times 10^{-31}\;\text{kg},$$
$$e = 1.60210 \times 10^{-19}\;\text{C},$$
and $$n$$ is the principal quantum number.

For the first orbit ($$n = 1$$) this radius is usually denoted by $$a_0$$ and called the Bohr radius. Setting $$n = 1$$ in the above formula, we have

$$ a_0 \;=\; \frac{\varepsilon_0 h^2}{\pi m e^2}. $$

Now we substitute all the given constants one by one:

$$ a_0 \;=\; \frac{(8.854185 \times 10^{-12}) \,(6.6262 \times 10^{-34})^2}{\pi \,(9.1091 \times 10^{-31})\,(1.60210 \times 10^{-19})^2}. $$

First, we square Planck’s constant:

$$ h^2 \;=\; (6.6262 \times 10^{-34})^2 \;=\; 43.893 \times 10^{-68} \;=\; 4.3893 \times 10^{-67}\;\text{J}^2\text{s}^2. $$

Next, we square the electronic charge:

$$ e^2 \;=\; (1.60210 \times 10^{-19})^2 \;=\; 2.5667 \times 10^{-38}\;\text{C}^2. $$

Now we multiply $$\varepsilon_0$$ and $$h^2$$ in the numerator:

$$ \varepsilon_0 h^2 \;=\; (8.854185 \times 10^{-12}) \times (4.3893 \times 10^{-67}) \;=\; 3.885 \times 10^{-78}. $$

In the denominator we have $$\pi m e^2$$, so we first multiply $$m$$ and $$e^2$$:

$$ m e^2 \;=\; (9.1091 \times 10^{-31}) \times (2.5667 \times 10^{-38}) \;=\; 2.337 \times 10^{-68}. $$

Then we multiply by $$\pi$$:

$$ \pi m e^2 \;=\; 3.1416 \times 2.337 \times 10^{-68} \;=\; 7.343 \times 10^{-68}. $$

Forming the ratio gives

$$ a_0 \;=\; \frac{3.885 \times 10^{-78}}{7.343 \times 10^{-68}} \;=\; 5.292 \times 10^{-11}\;\text{m}. $$

We recognize that $$1\;\text{Å} = 10^{-10}\;\text{m}$$, so

$$ a_0 \;=\; 5.292 \times 10^{-11}\;\text{m} \;=\; 0.529\;\text{Å}. $$

Thus the first Bohr radius is $$0.529$$ Å. For the second Bohr orbit we set $$n = 2$$ in the general formula. Because $$r_n = a_0 n^2$$, we have

$$ r_2 \;=\; a_0 (2)^2 \;=\; a_0 \times 4. $$

Substituting the value of $$a_0$$ we just calculated:

$$ r_2 \;=\; 4 \times 0.529\;\text{Å} \;=\; 2.116\;\text{Å}. $$

Rounding appropriately, we obtain

$$ r_2 \approx 2.12\;\text{Å}. $$

The option that matches this value is 2.12 Å.

Hence, the correct answer is Option C.