The total vapour pressure of a solution in this case only depends on vapour pressure of water as anhydrous oxalic acid is a non-volatile compound.
$\therefore$ Vapour pressure of solution = vapour pressure of
Water $\left(P_w\right)$
According to Raoult’s law
Number of moles of oxalic acid = $\frac{9}{90}=0.1$ moles
$\therefore x_w=\frac{9.9}{9.9+0.1}=0.99$
$\Rightarrow P_s=P_w=0.99\times P_1^o$