How many NaOH is required to neutralize 1500cm^3 of 0.1N HCl? a. 60g b. 6g c. 4g d. 40g

The answer is b) 6 g
Start by writing the balanced chemical equation for this
##NaOH_((aq)) + HCl_((aq)) -> NaCl_((aq)) + H_2O_((l))##
Notice that you have a ##1:1## between sodium hydroxide and hydrochloic acid; this means that a complete would require equal numbers of moles of sodium hydroxide and hydrochloric acid.
Now you know that the hydrochloric acid solution has a normality of 0.1 N. Normality is simply a measure of reactivity meaning that it is calculated by taking into account how a substance behaves in a particular reaction.
Hydrochloric acid dissociates in aqueous solution to rpoduce
##HCl_((aq)) -> H_((aq))^(+) + Cl_((aq))^(-)##
The net ionic equation for your reaction will be
##OH_((aq))^(-) + H_((aq))^(+) -> H_2O_((l))##
In your case a 0.1 N solution means that the hydrochloric acid solution provides 0.1 moles of protons ##H^(+)## per liter to the reaction.
Since 1 mole of ##HCl## produces 1 mole of ##H^(+)## the of the solution will be equal to ##0.1 mol/L##.
The volume of the solution will be
##1500color(red)(cancel(color(black)(cm^3))) * (1 dm^3)/(1000color(red)(cancel(color(black)(cm^3)))) = 1.5 dm^3##
This means that you can now calculate how many moles of hydrochloric acid took part in the reaction (remember that ##1 dm^3 = 1 L##)
##C = n/V implies n = C * V##
##n_HCl = 0.1 moles/color(red)(cancel(color(black)(L))) * 1.5color(red)(cancel(color(black)(L))) = 0.15 moles HCl##
The aforementioned mole ratio tells you that the number of moles of sodium hydroxide needed to neutralize this many moles of hydrochloric acid is
##n_(NaOH) = n_(HCl) = 0.15 moles##
To get the mass of sodium hydroxide needed use its molar mass
##0.15color(red)(cancel(color(black)(moles))) * 40.0 g/(1color(red)(cancel(color(black)(mole)))) = color(green)(6 g)##