The volume of water must be added to 100 mL of 0.5M NaOH solution to get 0.2M solution is.

Solution - I  (Initial or before dilution)                                                                 

Solution – II (Final after dilution)

Volume of  NaOH solution$={\text{V}}_1=100\text{ mL}$                        

Volume of NaOH $={\text{V}}_2$

${\text{Molarity,M}}_1=0.\text{5 M}$                                                      

${\text{Molarity,M}}_2=0.\text{2 M}$

According to Dilution Law,

        ${\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$

The final volume of NaOH is

${\text{V}}_2=\frac{{\text{M}}_1{\text{V}}_1}{{\text{M}}_2}=\frac{0.5\times 100}{0.2}=250\text{mL}$

Volume of water to be added $={\text{V}}_2 -{\text{ V}}_1$

                                             $=250-100$

                                              $=150\text{ mL}$