${H}_2O(s)\leftrightharpoons{H}_2O(l)$

A) the mass of ice must equal to the mass of water B) the mass water and the mass of ice each remain constant

At equilibrium, the amount of reactant being converted to product is equal to the amount of product being converted to reactant. Therefor, the concentration of both the reactant and product are no longer changing and the ratio of these no longer changing concentrations can be represented as a fixed number or constant - the equilibrium constant. $$\mathrm{K_{eq}=\frac{[Product]}{[Reactant]}}$$ where the brackets denote concentrations. In the case at hand, a liquid-solid equilibrium, the expression becomes (thanks to DavePhD for pointing this out, see his UC Davis link in the comments if you'd like more info.) $$\ce{K_{eq}=[H2O~ (l)]}$$ This equation tells us that at equilibrium, the concentration of liquid water is a constant, it is not changing. If the concentration of liquid water is not changing, then the concentration of ice cannot be changing. This does not say that the concentration (or mass, since concentration is related to mass divided by molecular weight per unit volume) of liquid water must equal the concentration or mass of ice; it just says that the concentrations or masses of liquid water and ice must be constant. Hence "B" is the correct choice.