5 % of 100 is 5. So we have to insert 5 elements for making the hash table 5% full.
Probability of collision during first insertion = 0
Probability of collision during second insertion = 1/100 (Collision occurs only if trying to insert in previously inserted location)
Probability of collision during third insertion = 2/100
Probability of collision during fourth insertion = 3/100
Here, it is asked to consider probability before 5% full. So we consider only 4 insertions. We need not consider fifth insertion, as it will make the table 5% full.
Therefore, total probabilty of collsion = 0 + (1/100) + (2/100) + (3/100) = 0.06 $\approx$ 0.1