how to prepare 100 ml of 0.4 N Naoh?

No. of Eq. of NaOH required = (0.4 Eq/L) × (100/1000 L) = 0.04 Eq
Equivalent mass of NaOH = (23.0 + 16.0 + 1.0) g/Eq = 40.0 g/Eq
Mass of NaOH required = (0.04 Eq) × (40.0 g/Eq) = 1.60 g

Procedure:
In a small beaker, dissolve 1.60 g of NaOH in a small amount of distilled water (or ionized water) to form a solution. Totally transfer the solution to a 100-ml volumetric flask, and then add distilled water (or ionized water) to the mark of the volumetric flask. Stopper and mix well the solution. The solution prepared is 100 ml of 0.4 N NaOH.

Since Na has a +1 charge, 0.4 N solution is the same as a 0.4 M solution. One liter of a 0.4 M solution contains 0.4 mole of NaOH. 100 ml is one tenth of liter. So 100 ml of this solution must contain 0.04 mole of NaOH.

NaOH = 12 + 1 + 16 = 40 grams per mole
0.04 * 40 = 1.6 grams

To prepare this solution, dissolve 1.6 grams of NaOH in 50 ml of demonized water. Stir. Then, add enough demonized water to have a total volume of 100 ml. I hope this is helpful for you.