The longest wavelength associated with Paschen series is : (Given $$R_H = 1.097 \times 10^7 \text{ SI unit}$$)

We need to find the longest wavelength associated with the Paschen series of hydrogen. The wavelengths of spectral lines in the hydrogen spectrum are given by the Rydberg formula: $$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$ where $$R_H = 1.097 \times 10^7 \,\text{m}^{-1}$$ is the Rydberg constant, $$n_1$$ is the lower energy level, and $$n_2$$ is the upper energy level.

For the Paschen series, transitions end at $$n_1 = 3$$, and the lines correspond to $$n_2 = 4, 5, 6, \ldots$$ The longest wavelength corresponds to the smallest energy transition, which is the transition from $$n_2 = 4$$ to $$n_1 = 3$$. Thus,

$$\frac{1}{\lambda} = R_H\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R_H\left(\frac{1}{9} - \frac{1}{16}\right)$$

Computing the difference of fractions gives

$$\frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144}$$

Substituting back into the Rydberg formula,

$$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{7}{144} = \frac{7.679 \times 10^7}{144} = \frac{7.679}{144} \times 10^7$$

$$\frac{1}{\lambda} = 0.05333 \times 10^7 = 5.333 \times 10^5 \,\text{m}^{-1}$$

Solving for $$\lambda$$ yields

$$\lambda = \frac{144}{7 \times 1.097 \times 10^7} = \frac{144}{7.679 \times 10^7} = 18.752 \times 10^{-7} \,\text{m} = 1.875 \times 10^{-6} \,\text{m}$$

Rounding gives $$1.876 \times 10^{-6}$$ m, so the correct answer is Option B: $$1.876 \times 10^{-6}$$ m.