Avoiding an accidental bottom's up

The first thing we need is the center of gravity of the empty glass. Let's make it relative to bottom of the inside of the glass.

The thick base volume is $\pi*3.55^{2}*1.4=55.43cm^{3}$ at a distance of $-0.7cm$

The cylindrical shell of the side is $\pi(3.55^{2}-3.35^{2})*12.6=54.63cm^{3}$ at a distance of $6.3cm$

Since the densities of these bits are the same we can find the center of gravity of the empty glass with a weighted mean

$\frac{55.43*-0.7+54.63*6.3}{55.43+54.63} = 2.77$

Call the depth of the water $x$ .

The water will add $\pi*3.35^{2}*x=35.26x grams$ at a height of $0.5x cm$ . We can now find the cog as a function of $x$ by another weighted mean:

$C(x) = \frac{2.77*275+0.5x*35.26x}{275+35.26x}$

$C(x)= \frac{761.75+17.63x^{2}}{275+35.26x}$

This will be minimized when the water level is the same as the cog (can you see why?)

$\frac{761.75+17.63x^{2})}{275+35.26x}=x$

$761.75+17.63x^{2}=275x+35.26x^{2}$

$0=17.63x^{2}+275x-761.75$

The quadratic equation has positive solution $x=2.40$

Which corresponds to a mass of $35.26*2.4 = 84.624g$ or volume of $\boxed{85 } cm^{3}$