What is the molality of a dilute aqueous 0.02N $H_{3}P{O}_4$ solution?

The correct option is D 0.0067
Given,
Normality, N = 0.02
It is known that $N=n_1\times M$, where $n$ is the n-factor of equivalence and M is the molarity of the solution.
On expanding, $N=n_1\times \frac{\text{Number of moles}}{\text{Volume of solution(L)}}$
For $H_{3}P{O}_4$, $n=3$ since there are 3 breakable OH bonds.
Therefore, $N=3\times \frac{\text{Number of moles}}{\text{Volume of solution(L)}}$. Here, volume of solution (1 L) can be approximated as mass of solvent (1 kg) since it is an dilute solution in water (density = 1 kg/L) implying water concentration is far greater such that amount of $H_{3}P{O}_4$ is negligible.
Now, $\frac{\text{Number of moles}}{\text{Volume of solution(kg)}}=\text{molality of solution}$
Thus, $\frac{N}{3}=0.02/3=0.0067molal.$