No 'accept' after 10 hrs for the very nice answer by @Mick A. --Among 'inattentive', 'unappreciative', and 'confused', I'll assume confused. (New users do
not always understand that relevant follow-up questions are OK, or realize that
clicking to 'accept' takes the question off our list of ones without satisfactory answers.)
From the incorrect
speculation in the question that the answer is 5+5+5, I think there may
be confusion about the distribution of the waiting time for the first
customer to arrive. That distribution is exponential with rate $\lambda$, hence mean $1/\lambda$ and variance $1/\lambda^2$. So the variance of the waiting
time for the first customer is 1/25; and then the variance of the waiting time for the third customer is 1/25 + 1/25 + 1/25.
Notes: (1) Eventually, in doing more with queues, it is essential to know that the name of the distribution of the waiting time for the $n$th customer is the 'gamma' (here also 'Erlang') distribution, but that information is not necessary to solve the elementary problem asked. (2) A Poisson random variable has mean numerically equal to variance; an exponential random variable has mean equal to standard deviation.