Energy of an electron is given by $$E = -2.178 \times 10^{-18}\left(\frac{Z^2}{n^2}\right)$$ J. Wavelength of light required to excite an electron in a hydrogen atom from level n=1 to n=2 will be :
$$(h = 6.62 \times 10^{-34}$$ Js and $$c = 3.0 \times 10^8$$ ms$$^{-1})$$

The energy of an electron in a hydrogen-like atom is given in the statement as

$$E = -2.178 \times 10^{-18}\left(\frac{Z^{2}}{n^{2}}\right)\,\text{J}.$$

For the hydrogen atom we have $$Z = 1$$, so the formula simplifies to

$$E = -2.178 \times 10^{-18}\left(\frac{1}{n^{2}}\right)\,\text{J}.$$

The electron is to be promoted from the ground level $$n = 1$$ to the first excited level $$n = 2$$. Let us calculate the energy of each level separately.

For $$n = 1$$:

$$E_{1} = -2.178 \times 10^{-18}\left(\frac{1}{1^{2}}\right) = -2.178 \times 10^{-18}\ \text{J}.$$

For $$n = 2$$:

$$E_{2} = -2.178 \times 10^{-18}\left(\frac{1}{2^{2}}\right) = -2.178 \times 10^{-18}\left(\frac{1}{4}\right) = -\frac{2.178}{4} \times 10^{-18}\ \text{J} = -5.445 \times 10^{-19}\ \text{J}.$$

The energy that must be absorbed is the difference between the two levels:

$$\Delta E = E_{2} - E_{1}.$$

Substituting the values we have just obtained,

$$\Delta E = \left(-5.445 \times 10^{-19}\right) - \left(-2.178 \times 10^{-18}\right)$$ $$\Delta E = -5.445 \times 10^{-19} + 2.178 \times 10^{-18}$$ $$\Delta E = \left(2.178 - 0.5445\right) \times 10^{-18}$$ $$\Delta E = 1.6335 \times 10^{-18}\ \text{J}.$$

(A more rounded value, $$1.634 \times 10^{-18}\ \text{J}$$, will give the same final wavelength.)

According to the Planck-Einstein relation, the energy of a photon is connected to its frequency by

$$\Delta E = h\nu,$$

and the frequency and wavelength are related through the speed of light:

$$\nu = \frac{c}{\lambda}.$$

Combining the two relations gives the wavelength directly:

$$\Delta E = h\frac{c}{\lambda}\quad\Longrightarrow\quad \lambda = \frac{hc}{\Delta E}.$$

Now substitute the given constants $$h = 6.62 \times 10^{-34}\ \text{J\,s}$$ and $$c = 3.0 \times 10^{8}\ \text{m\,s}^{-1}$$ together with $$\Delta E = 1.6335 \times 10^{-18}\ \text{J}$$:

$$\lambda = \frac{\left(6.62 \times 10^{-34}\right)\left(3.0 \times 10^{8}\right)}{1.6335 \times 10^{-18}}.$$

Multiplying the numerator:

$$6.62 \times 3.0 = 19.86,$$ so $$h\,c = 19.86 \times 10^{-26}\ \text{J\,m} = 1.986 \times 10^{-25}\ \text{J\,m}.$$

Now divide:

$$\lambda = \frac{1.986 \times 10^{-25}}{1.6335 \times 10^{-18}} = 1.216 \times 10^{-7}\ \text{m}.$$

This is essentially the same as $$1.214 \times 10^{-7}\ \text{m}$$ when rounded to three significant figures, which matches one of the options.

Hence, the correct answer is Option C.