The weight of the hydrometer = weight of vol of liquid displaced (Archimedes principle). Let m be the mass , L be the length , and A be the cross sectional area of the hydrometer. Let d1, d2 , d3 be the densities of the liquids and x1, x2, x3 be the length of the stem above the water level . We have
mg = d1(Lx1) A g = d2(Lx2) A g = d3(Lx3) A g.
Note that the area cancels out. Now plug in the values and solve.
I got L = 14 and x3 = 5.111 cm
