Take 3min as a time unit. The mean for incoming calls is 1 (because a totally 20 calls per hour). So, the question is now:

Given a Poisson process with incoming rate of 1 in every 3min (mean =1), what is the probability that in any given 3min window, there is more than 3 incoming calls?

P(x>=4) = 1-P(0)-P(1)-P(2)-P(3)
= 1- 0.98 =0.02.

So there is a 0.02 chance a customer gets delayed.

I am not sure if my answer is correct. I would like to hear from professional statisticians who bump into this post.