Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be:

For any transition in the hydrogen atom we begin with the Rydberg formula for wavelength:

$$\frac1\lambda = R\left(\frac1{n_f^{\,2}} - \frac1{n_i^{\,2}}\right)$$

where $$R$$ is the Rydberg constant, $$n_i$$ is the initial (higher) quantum number and $$n_f$$ is the final (lower) quantum number. In the Balmer series we always have $$n_f = 2$$.

The data given tell us that for the first Balmer line, i.e. the transition $$n_i = 3 \to n_f = 2$$, the wavelength is

$$\lambda_1 = 660\ \text{nm}.$$

Using the formula, the reciprocal of this wavelength is

$$\frac1{\lambda_1}=R\!\left(\frac1{2^{2}}-\frac1{3^{2}}\right).$$

For the second Balmer line we have the transition $$n_i = 4 \to n_f = 2$$, so its wavelength $$\lambda_2$$ satisfies

$$\frac1{\lambda_2}=R\!\left(\frac1{2^{2}}-\frac1{4^{2}}\right).$$

To eliminate the unknown constant $$R$$ we take the ratio of the two reciprocal wavelengths:

$$\frac{\dfrac1{\lambda_2}}{\dfrac1{\lambda_1}} =\frac{\displaystyle R\left(\dfrac1{2^{2}}-\dfrac1{4^{2}}\right)} {\displaystyle R\left(\dfrac1{2^{2}}-\dfrac1{3^{2}}\right)} =\frac{\dfrac1{2^{2}}-\dfrac1{4^{2}}} {\dfrac1{2^{2}}-\dfrac1{3^{2}}}.$$

Because the common factor $$R$$ cancels, we invert the ratio to relate the wavelengths directly:

$$\frac{\lambda_2}{\lambda_1} =\frac{\dfrac1{2^{2}}-\dfrac1{3^{2}}} {\dfrac1{2^{2}}-\dfrac1{4^{2}}}.$$

Now we evaluate each term carefully, showing every fraction:

$$\frac1{2^{2}}=\frac14,\qquad \frac1{3^{2}}=\frac19,\qquad \frac1{4^{2}}=\frac1{16}.$$

So

$$\frac1{2^{2}}-\frac1{3^{2}} =\frac14-\frac19 =\frac{9-4}{36} =\frac5{36},$$

and

$$\frac1{2^{2}}-\frac1{4^{2}} =\frac14-\frac1{16} =\frac{4-1}{16} =\frac3{16}.$$

Substituting these exact fractions into the ratio gives

$$\frac{\lambda_2}{\lambda_1} =\frac{\dfrac5{36}}{\dfrac3{16}} =\frac5{36}\times\frac{16}{3} =\frac{5\times16}{36\times3} =\frac{80}{108} =\frac{20}{27}.$$

Now we multiply by the known value of $$\lambda_1$$ to find $$\lambda_2$$:

$$\lambda_2 =\lambda_1\left(\frac{20}{27}\right) =660\ \text{nm}\times\frac{20}{27}.$$

Carrying out the multiplication step by step, first divide 660 by 27:

$$\frac{660}{27}=24.444\ldots$$

and then multiply by 20:

$$24.444\ldots\times20=488.888\ldots\ \text{nm}.$$

Rounding to one decimal place we obtain

$$\lambda_2 \approx 488.9\ \text{nm}.$$

Hence, the correct answer is Option B.