How many times $0.2\mathrm{M}$ solution of sodium Acetate be diluted at $25^{\circ }\mathrm{C}$ to double the degree of hydrolysis?

The degree of hydrolysis for salt is given as:
${\mathrm{K}}_{\mathrm{h}}={\mathrm{h}}^2\mathrm{C}$
Where $K_h$ is the equilibrium constant for the hydrolysis reaction
' $h$ ' is the degree of hydrolysis of the reaction
'$\mathrm{C}$ ' is the concentration of the solution.
Let consider the concentration of sodium acetate as ${\mathrm{C}}_1$ then the equilibrium constant for the sodium acetate is written as:
${\mathrm{K}}_{\mathrm{h}}={\mathbf{h}}^2{\mathrm{C}}_1$
Here, ${\mathrm{C}}_1=0.2\mathrm{M}$ and consider ' $h$ ' as the degree of hydrolysis. Then, we have
${\mathrm{K}}_{\mathrm{h}}=(\mathrm{h}{)}^2(0.2)$..........(1).
When the degree of hydrolysis is doubled, the concentration would be,
${\mathrm{K}}_{\mathrm{h}}=(2\mathrm{h}{)}^2{\mathrm{C}}_2$..........(2).
Divide the equation (1) by equation (2). We get
$\begin{array}{rl}\frac{{\mathrm{K}}_{\mathrm{h}}}{{\mathrm{K}}_{\mathrm{h}}}& =\frac{(\mathrm{h}{)}^2(0.2)}{(2\mathrm{h}{)}^2{\mathrm{C}}_2}\\ 1& =\frac{0.2}{4{\mathrm{C}}_2}\\ {\mathrm{C}}_2& =0.05\mathrm{M}\end{array}$
The concentration of solution when we double the degree of hydrolysis is equal to 0.05 M
$\begin{array}{rl}{\mathrm{M}}_1{\mathrm{V}}_1& ={\mathrm{M}}_2{\mathrm{V}}_2\\ (0.2)\left({\mathrm{V}}_1\right)& =(0.05){\mathrm{V}}_2\\ {\mathrm{V}}_2& =\frac{0.2}{0.05}{\mathrm{V}}_1\\ \therefore {\mathrm{V}}_2& =4{\mathrm{V}}_1\end{array}$
Thus, the volume of the 0.2 M sodium acetate solution should be four times the initial volume so that the degree of hydrolysis would be double than the initial degree of hydrolysis.