Answer
$V=270.3\ L$
Work Step by Step
From item 1, there are $4.092\times10^{-3}\ mols$ of $CO_2$ in the headspace, therefore:
$3.7\times10^{-4}\ atm\times V=4.092\times10^{-3}\ mol\times0.082\ atm.L/mol.K\times (273+25)K$
$V=270.3\ L$
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