Answer
$3.66\times 10^5\;\rm m/s$
Work Step by Step
Our system here is isolated before and after the decay process. This means that the momentum is conserved.
The nucleus of the polonium isotope decays to an alpha particle plus a daughter nucleus.
$$p_{ix}=p_{fx}$$
$$m_pv_{ix,p}=m_{He}v_{fx,He}+m_Nv_{fx,d}$$
where $p$ refers to the polonium and $d$ refers to the daughter nucleus, and $He$ refers to alpha particle.
The initial speed of the polonium is zero since it was at rest before decaying.
$$0=m_{He}v_{fx,He}+m_dv_{fx,d}$$
Solving for $v_{fx,d}$;
$$v_{fx,d}=\dfrac{-m_{He}v_{fx,He}}{m_d}$$
The negative sign indicates that the velocity of the daughter nucleus is in the opposite direction of the alpha particle
Plugging the known and recall that the mass of the daughter nucleus is given by $m_d=m_p-m_{He}=214-4=210\;\rm u$;
$$v_{fx,d}=\dfrac{-4(1.92\times 10^7)}{210}=-\color{red}{\bf 3.66\times 10^5}\;\rm m/s$$
