Let's break down the problem step by step:

In the photoelectric effect, the energy of a photon is given by:

$$ h\nu = \phi + K_e $$

where:

$$\phi$$ is the work function of the metal,

$$K_e$$ is the kinetic energy of the ejected electron.

The kinetic energy of the ejected electrons is provided by the stopping potential:

$$ K_e = eV $$

Given that the stopping potential is $$ V = 2 \text{ V} $$, we have:

$$ K_e = 2 \text{ eV} $$

Now, we can calculate the photon energy:

$$ h\nu = \phi + K_e = 2.14 \text{ eV} + 2 \text{ eV} = 4.14 \text{ eV} $$

The relation between the photon's energy and its wavelength is:

$$ h\nu = \frac{hc}{\lambda} $$

Rearranging for $$\lambda$$:

$$ \lambda = \frac{hc}{h\nu} $$

Given that $$ hc = 1242 \text{ eV·nm} $$, substitute the values:

$$ \lambda = \frac{1242 \text{ eV·nm}}{4.14 \text{ eV}} $$

Calculate the wavelength:

$$ \lambda \approx \frac{1242}{4.14} \approx 300 \text{ nm} $$

Thus, the wavelength of the incident electromagnetic wave is approximately 300 nm.

The correct answer is Option C: 300 nm.