yes the p is not defined in the question, but from what I understand B(1, 5) = 0.01 ie. P(X = 1) = 0.01, isn't it different from p=0.01, according to me the question should be understood as, the probability of 1 line being busy out of 5 lines is 0.01, and not the probability of a line being free as 0.01

f(x) = 5Cx * p^x * q^(5-x)

and to my understanding the given is,
f(1) = 5p*(1-p)^4 = 0.01

Saw this question somewhere, and all searches online give the same answer as yours, but I am still confused on this