Answer
$V_{spill out}=0.26 cm^3$
Work Step by Step
Co-efficient of volume expansion for aluminium cup
$\gamma_{Al}=3(\alpha_{Al})=3(23\times 10^{-6})=6.9\times10^{-5}K^{-1}$
so increase in the volume of cup is
$\Delta{V_{Al}}=\gamma_{Al}V_{\circ}\Delta{T}$
put the values
$\Delta{V_{Al}}=(6.9\times10^{-5})(100\times10^{-6})(6)$
$\Delta{V_{Al}}=4.14\times10^{-8}m^3$
Now we find the increase in the volume of the glycerin
$\Delta{V_{glycerin}}=\gamma_{glycerin}V_{\circ}\Delta{T}=5.1\times10^{-4}(100\times10^{-6})(6)$
$\Delta{V_{glycerin}}=3.06\times10^{-7} m^3$
For the volume of the glycerin that spills out
$V_{spill out}=\Delta{V_{glycerin}}-\Delta{V_{Al}}$
putting the values, we get
$V_{spill out}= 3.06\times10^{-7}-4.14\times10^{-8}$
$V_{spill out}=2.65\times10^{-7}m^3$
or
$V_{spill out}=0.26 cm^3$
