Answer
$3.69L$
Work Step by Step
We can calculate the required volume as follows:
$P_{He}+P_{H_2O}=1.00atm=760torr$
$\implies P_{He}+23.8torr=760torr$
$\implies P_{He}=736torr$
and number of moles of $H_e$ can be calculated as
$n_{He}=0.586g\times \frac{1mol}{4.003g}=0.146mol\space He$
Now $V=\frac{n_{He}RT}{P_{He}}$
We plug in the known values to obtain:
$V=\frac{0.146mol\times \frac{0.08206L\space atm\times 298K}{K\space mol}}{736torr\times \frac{1atm}{760torr}}=3.69L$
