Answer
$0.0214\;\rm m/s$
Work Step by Step
We have here an isolated system [the bullet+the target] since the surface is frictionless and since there are no external forces exerted on our system horizontally. This means that the momentum is conserved.
We assume that the target will not move until the bullet comes out from the other side since it is too heavy relative to the bullet.
$$p_{ix}=p_{fx}$$
$$m_bv_{b,ix}+m_tv_{t,ix}=m_bv_{b,fx}+m_tv_{f,ix}$$
the heavy target was initially at rest.
$$m_bv_{b,ix}+0=m_bv_{b,fx}+m_tv_{t,fx}$$
Solving for $v_{t,fx}$;
$$v_{t,fx}=\dfrac{m_bv_{b,ix}-m_bv_{b,fx}}{m_t} $$
Plugging the known;
$$v_{t,fx}=\dfrac{(0.025\times 1200)-(0.025\times 900)}{350} $$
$$v_{t,fx}=\color{red}{\bf 0.0214}\;\rm m/s$$
