Identify the formula for the energy of a photon.

The energy of an emitted photon ($E$) is given by the formula:

$$E = \frac{hc}{\lambda}$$

where $h$ is Planck's constant, $c$ is the speed of light (emitted radiation speed), and $\lambda$ is the wavelength of the photon.

Given:
Planck's constant, $h = 6.63 \times 10^{-34} \text{ Js}$
Speed of emitted radiation, $c = 3.0 \times 10^8 \text{ ms}^{-1}$
Wavelength of emitted photon, $\lambda = 5.68 \times 10^{-6} \text{ m}$

$$E = \frac{(6.63 \times 10^{-34} \text{ Js}) \times (3.0 \times 10^8 \text{ ms}^{-1})}{5.68 \times 10^{-6} \text{ m}}$$

First, calculate the numerator:

$$6.63 \times 3.0 = 19.89$$ $$10^{-34} \times 10^8 = 10^{(-34+8)} = 10^{-26}$$

So, the numerator is $19.89 \times 10^{-26} \text{ Jm}$.

Now, divide by the wavelength:

$$E = \frac{19.89 \times 10^{-26} \text{ Jm}}{5.68 \times 10^{-6} \text{ m}}$$ $$E = \left(\frac{19.89}{5.68}\right) \times \left(\frac{10^{-26}}{10^{-6}}\right) \text{ J}$$ $$E \approx 3.50176 \times 10^{(-26 - (-6))} \text{ J}$$ $$E \approx 3.50176 \times 10^{-20} \text{ J}$$

Rounding to two decimal places for the coefficient, we get $3.50 \times 10^{-20} \text{ J}$.
Comparing this to the options:
A $1.49 \times 10^{-20} \text{ J}$
B $1.49 \times 10^{-19} \text{ J}$
C $3.49 \times 10^{-20} \text{ J}$
D $3.49 \times 10^{-19} \text{ J}$

The calculated value $3.50 \times 10^{-20} \text{ J}$ is closest to option C, $3.49 \times 10^{-20} \text{ J}$.

The final answer is $\text{C}$.