Answer
$8.022 atm$
Work Step by Step
We can calculate the number of moles for $O_2$ as
$n=91.3\times \frac{1 mol\space O_2}{32.00g\space O_2}=2.853mol O_2$
Now according to ideal gas law
$PV=nRT$
$\implies P=\frac{nRT}{V}$
We plug in the known values to obtain:
$P=\frac{2.853\times 0.08206\times 294}{8.58}=8.022atm$