To find the equivalent focal length of the lens combination, we use the lens maker's formula.$\\$Given:$\\ n_1 = 1.5, \\ n_2 = 1.6, \\ R_1 = R_2 = 20 \,$cm$\\$The lens is placed in a medium with refractive index $n_2. \\$The lens maker's formula for a lens in a medium is:$\\ \frac{1}{f} = \left(\frac{n_1}{n_2} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \\$Substitute the given values:$\\ \frac{1}{f} = \left(\frac{1.5}{1.6} - 1\right) \left(\frac{1}{20} - \frac{1}{-20}\right) \\$Simplify:$\\ \frac{1}{f} = \left(\frac{1.5}{1.6} - 1\right) \left(\frac{1}{20} + \frac{1}{20}\right) \\ \frac{1}{f} = \left(\frac{1.5 - 1.6}{1.6}\right) \left(\frac{2}{20}\right) \\ \frac{1}{f} = \left(\frac{-0.1}{1.6}\right) \left(\frac{1}{10}\right) \\ \frac{1}{f} = \frac{-0.1}{16} \\ f = -160 \,$cm$\\$Therefore, the equivalent focal length of the lens combination is $-100 \,$cm$. \\$This corresponds to Option 3.