Answer
$4.82\times 10^5\;\rm m/s$
Work Step by Step
We can consider the proton plus the golden target as an isolated system. So, the momentum is conserved.
$$p_{ix}=p_{fx}$$
$$m_pv_{ix,p}+m_g\overbrace{v_{ix,g}}^{=0}=m_pv_{fx,p}+m_gv_{fx,g}$$
where $p$ refers to the proton and $g$ refers to the golden target. The initial speed of the golden target is zero since it was at rest.
Now we need to solve for $v_{fx,g}$;
$$m_pv_{ix,p} -m_pv_{fx,p}=m_gv_{fx,g}$$
$$ v_{fx,g}=\dfrac{m_pv_{ix,p} -m_pv_{fx,p}}{m_g} $$
We know that the recoil speed of the proton is 0.9 its initial speed but in the opposite direction. so $v_{fx,p}=-0.9v_{ix,p}$
$$ v_{fx,g}=\dfrac{m_pv_{ix,p} -(-0.9m_pv_{ix,p})}{m_g} $$
$$ v_{fx,g}=\dfrac{1.9m_pv_{ix,p} }{m_g} $$
Plugging the known;
$$ v_{fx,g}=\dfrac{ 1.9\times 1 \times 5.0\times 10^7 }{197} =\color{red}{\bf 4.82\times 10^5}\;\rm m/s $$
