Probability of item being defective = $p = \frac{20}{100} = \frac{1}{5}$

Probability that less than 2 items are defective = Prob(no item is defective) + Prob(1 item is defective). Hence required probability is

$$P = \binom{4}{0}\left(\frac{1}{5}\right)^0\left(\frac{4}{5}\right)^4 + \binom{4}{1}\left(\frac{1}{5}\right)^1\left(\frac{4}{5}\right)^3 = \frac{512}{625}$$

Hence option (4) is correct.