Answer
$6.588$ mL of water can be added to the glass jar.
Work Step by Step
We first calculate the change in volume of the glass due to thermal expansion.
$$\Delta V=bV_{0}\Delta T $$ $$V_{f}-450mL=(27\times10^{-6})(450mL)(20.0^{\circ}C-100^{\circ}C)$$ $$V_{f}=449.028 mL$$
Now, we calculate the change in volume of the water due to thermal expansion.
$$\Delta V=bV_{0}\Delta T $$ $$V_{f}-450mL=(210\times10^{-6})(450mL)(20.0^{\circ}C-100^{\circ}C)$$ $$V_{f}=442.44 mL$$
Therefore, the amount of water that could be added is $449.028 mL-442.44 mL=6.588mL$.