The limiting line in Balmer series will have a frequency of (Rydberg constant, $$R_\infty = 3.29 \times 10^{15}\ \text{cycles/s}$$)

The Balmer series corresponds to electronic transitions that terminate at the level $$n_1 = 2$$ in the hydrogen atom.

For any line of a hydrogen series, the Rydberg formula gives the frequency $$\nu$$ as:
$$\nu = R_\infty\left(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right)$$ where $$n_1$$ is the lower (final) level and $$n_2$$ (>$$n_1$$) is the upper (initial) level.

The limiting (or series) line is obtained when the electron falls from $$n_2 = \infty$$ to $$n_1 = 2$$. At $$n_2 = \infty$$, the term $$\dfrac{1}{n_2^{2}}$$ becomes zero. Hence
$$\nu_{\text{limit}} = R_\infty\left(\frac{1}{2^{2}} - 0\right) = R_\infty \times \frac{1}{4}$$

Insert the given value $$R_\infty = 3.29 \times 10^{15}\ \text{cycles s}^{-1}$$:
$$\nu_{\text{limit}} = \frac{3.29 \times 10^{15}}{4} = 0.8225 \times 10^{15}\ \text{s}^{-1} = 8.225 \times 10^{14}\ \text{s}^{-1}$$

Rounded to three significant figures, $$\nu_{\text{limit}} \approx 8.22 \times 10^{14}\ \text{s}^{-1}$$.

Option A which is: $$8.22 \times 10^{14}\ \text{s}^{-1}$$