$$\begin{gathered} {\mathrm{\pi }}_1={\mathrm{C}}_1{\mathrm{ST}}_1; \\ {\mathrm{\pi }}_2={\mathrm{C}}_2{\mathrm{ST}}_2; \\ \frac{{\mathrm{\pi }}_1}{{\mathrm{\pi }}_2}=\frac{{\mathrm{C}}_1{\mathrm{T}}_1}{{\mathrm{C}}_2{\mathrm{T}}_2} \end{gathered}$$

For initial solution, ${\mathrm{\pi }}_1=\frac{400}{760}$ atm, ${\mathrm{T}}_1=273\mathrm{K}$

After dilution, let volume becomes ${\mathrm{V}}_2$ and temperature is raised to $20°\mathrm{C},$ i.e., ${\mathrm{T}}_2=293 \mathrm{K}. \mathrm{and} {\mathrm{\pi }}_2 = \frac{100}{760} \mathrm{atm}$

On putting the values we get

$$\begin{gathered} \frac{400}{100} = \frac{C_1}{C_2}\times \frac{273}{293} \\ \frac{{\displaystyle C_1}}{{\displaystyle C_2}} = 4.3 \end{gathered}$$

i.e., the final solution is $4.3$ times more diluted than the initial solution.