Molality $=\frac{ \mathrm{Mass} \mathrm{of} \mathrm{solute} \times 1000}{ \mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{solute} \times \mathrm{Mass} \mathrm{of} \mathrm{solvent} }$

Mass of solvent 

$=250 \mathrm{mL}\times 19 \mathrm{g}/\mathrm{mL}=475 \mathrm{g}$

$\therefore$ Mass of solvent $=472 \mathrm{g}$

Molar mass of ${\left({\mathrm{CO}}_2\mathrm{H}\right)}_2.2{\mathrm{H}}_2\mathrm{O}=126 \mathrm{g} {\mathrm{mol}}^{-1}$

$\therefore$ Molality $=\frac{3\times 1000}{472\times 126}$

$=0.05 \mathrm{mol} {\mathrm{kg}}^{-1}$

Normality $=\frac{ \mathrm{Number} \mathrm{of} \mathrm{equivalents} \mathrm{of} \mathrm{solute} }{ \mathrm{Volume} \mathrm{of} \mathrm{solution} \left(l\right)}$

$=\frac{ \mathrm{Mass} \mathrm{of} \mathrm{solute} \left(\mathrm{g}\right)\times 1000}{ \mathrm{Equivalent} \mathrm{mass} \mathrm{of} \mathrm{solute} }\times$ Volume of solution $\left(\mathrm{mL}\right)$

Equivalent mass of oxalic acid

$=\frac{\mathrm{Molar}mass}{2}$

$=63 \mathrm{g}/\mathrm{equi}$.

$\therefore$ Normality $=\frac{3\times 1000}{63\times 250}$

$=0.19$ (equivalents)$/\mathrm{l}$

or $0.19 \mathrm{N}$