A problem on Equivalent focal length

Consider the focal length of the simplest lens:

You can see that if you send parallel rays from 2 to 1, the focal length is $f_1=frac{n_1R}{n_2-n_1}$. But if you send rays in opposite direction, the focal length is $f_2=frac{n_2R}{n_2-n_1}$. That's why it's more beneficial to talk about surface power (optical power which doesn't depend on on what side of the lens we are): $$ S = frac{n}{f_n} = frac{n_2-n_1}R = nP_n. $$ For scenarios when the media before and after the lens is the same, the Lens Maker's formula of the sum of surface powers becomes the sum of optical powers: $$ P = frac{S}{n} = frac{S_1}n+frac{S_2}{n} = P_1 + P_2 $$

One way to avoid it is to decompose the lenses to have the same outside medium, usually air. This what is suggested in your problem: you have a glass lens in the air, a water lens in the air, and one more glass lens in the air. The sum of their optical powers: $$ P = 2frac{n_g-n_0}{Rn_0} + 2frac{n_w-n_0}{-Rn_0} + 2frac{n_g-n_0}{Rn_0} = frac{4n_g-2n_w-2n_0}{Rn_0}. $$

However, if you know about surface powers, you can avoid virtual air between water and glass: $$ n_0P = S = frac{n_g-n_0}{R} + frac{n_w-n_g}{-R} + frac{n_g-n_w}{R} + frac{n_0-n_g}{-R} = frac{4n_g-2n_w-2n_0}{R} $$