Given:

Volume, V = 500cm\(^3\)

Concentration , C = 0.2 M

n = \(\frac{CV}{1000}\)

n = \(\frac{{0.2}\times{500}}{1000}\)

n = 0.1 mole

Also, recall that n = \(\frac{mass}{Molar mass}\) = \(\frac{m}{M}\)

  m = n X M 

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

  m = 0.1 x  40 = 4g