Answer
An almost identical problem is done on page 101. Let v represent the little fish's velocity before lunch.
Set the momentum before lunch equal to the momentum after.
$$(5 kg)(1 m/s) + (1 kg) v = 0$$
$$5 m/s + v = 0$$
$$v = -5 m/s$$
So the little fish was approaching the big fish at at speed of 5 m/s. Its initial momentum was equal and opposite to that of the big fish, because the system's momentum was zero.