Molar mass of oxalic acid = $126 \mathrm{g}/\mathrm{mol}$

Mole =$\frac{\mathrm{Given} \mathrm{mass}}{\mathrm{Molar} \mathrm{mass}}=\frac{0.63}{126}=0.005 \mathrm{mol}$

Molarity = $\frac{\mathrm{Mole} \mathrm{of} \mathrm{solute}}{\mathrm{Volume} \mathrm{of} \mathrm{solution} \mathrm{in} \mathrm{ml}}\times 1000=\frac{0.005}{250}\times 1000=0.02\mathrm{M}$

Equivalent mass of oxalic acid = $\frac{126}{2}=63$

Gram equivalent of the solute = $\frac{0.63}{63}=0.01$

$\mathrm{Normality}=\frac{\mathrm{Gram} \mathrm{equivalent} \mathrm{of} \mathrm{solute}}{\mathrm{Volume} \mathrm{of} \mathrm{solution} \mathrm{in} \mathrm{ml}}\times 1000=\frac{0.01}{250}\times 1000=0.04\mathrm{N}$

Hence, normality of solution is $0.04 \mathrm{N}$.