To find the focal length of a lens with a refractive index of 1.6 when placed in water, given that its focal length in air is 12 cm and the refractive index of water is 1.28, we can use the lens maker's formula:

$ \frac{1}{f} = \left(\frac{\mu_L}{\mu_m} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

When the lens is in air, the refractive index of the medium, $\mu_m$, is 1. Therefore, for the lens in air:

$ \frac{1}{12} = (1.6 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

Simplifying gives:

$ \frac{1}{12} = \frac{6}{10} \left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

Solving for $\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$:

$ \frac{1}{R_1} - \frac{1}{R_2} = \frac{10}{72} $

Now, when the lens is placed in water, $\mu_m = 1.28$. Substituting into the equation gives:

$ \frac{1}{f} = \left(\frac{1.6}{1.28} - 1\right) \left(\frac{10}{72}\right) $

Calculating each part:

$ \frac{1}{f} = \frac{32}{128} \times \frac{10}{72} $

$ \frac{1}{f} = \frac{1}{4} \times \frac{10}{72} $

Hence, solving for $f$:

$ f = 28.8 \, \text{cm} = 288 \, \text{mm} $

Thus, the focal length of the lens in water is 288 mm.