Answer
The volume that overflowed is the subtraction of the expansions:
$\Delta V _Hg -\Delta V _g =\beta _Hg V_0 \Delta T - \beta _g V_0 \Delta T $
Now it is necessary to isolate $ \beta _g$:
$\Delta V _Hg -\Delta V _g = V_0 \Delta T ( \beta _Hg - \beta _g)$
$\frac{\Delta V _Hg -\Delta V _g}{V_0 \Delta T} = \beta _Hg - \beta _g$
$\beta _g = \beta _Hg - \frac{\Delta V _Hg -\Delta V _g}{V_0 \Delta T}$
Evaluating:
$ \beta _g = 18.0 \times 10 ^{-5} K^{-1} - \frac{8.95(m^{-2})^{3}}{55K\times 1000(m^{-2})^{3}} = 1.7 \times 10^{-5} K^{-1}$
Work Step by Step
First, knowing that the volume that overflowed was equal to the subtraction of the expansions, I wrote this and replaced it with the expression for the thermal expansion. After that I just isolated the desired variable and evaluated the equation.
