Answer
$-31.5\;\rm m/s$
Work Step by Step
Since the drag force is negligible during the small time interval of expelling the water, the system [Squids+ejected water] is isolated.
So that the momentum is conserved and hence,
$$p_{ix}=p_{fx}$$
The Squids plus ejected water are initially one unit, so
$$(m_w+m_s)v_{ix}=m_wv_{fx,w}+m_sv_{fx,s}$$
where $w$ refers to the ejected water, and $s$ refers to the squids.
Solving for $v_{fx,w}$;
$$v_{fx,w} =\dfrac{(m_w+m_s)v_{ix} -m_sv_{fx,s} }{m_w}$$
Plugging the known;
$$v_{fx,w} =\dfrac{[(0.1+1.5)\times 0.40 ] -(1.5\times2.5) }{0.1}$$
$$v_{fx,w} =\bf -31.1\;\rm m/s$$
Thus, the speed of the ejected water relative to the squids is given by
$$v'_{fx,w} =v_{fx,w} -v_{ix,s} = -31.1-0.40 =\color{red}{\bf -31.5}\;\rm m/s $$
