Answer
$14\;\rm u$
Work Step by Step
We can consider the proton plus the golden target as an isolated system. So, the momentum is conserved.
$$p_{ix}=p_{fx}$$
$$m_pv_{ix,p}+m_t\overbrace{v_{ix,t}}^{=0}=m_pv_{fx,p}+m_tv_{fx,t}$$
where $p$ refers to the proton and $t$ refers to the target. The initial speed of the target is zero since it was at rest.
Now we need to solve for $m_{t}$;
$$m_pv_{ix,p} -m_pv_{fx,p}=m_tv_{fx,t}$$
$$m_t =\dfrac{m_pv_{ix,p} -m_pv_{fx,p}}{v_{fx,t}} $$
We know that the recoil speed of the proton is reduced by 25%, so its final speed is now about 0.75 its initial speed but in the opposite direction. so $v_{fx,p}=-0.75v_{ix,p}$
$$ m_t =\dfrac{m_pv_{ix,p} -(-0.75m_pv_{ix,p})}{v_{fx,t}} $$
$$ m_t =\dfrac{1.75m_pv_{ix,p}}{v_{fx,t}} $$
Plugging the known;
$$ m_t =\dfrac{1.75(1)( 2.5\times 10^6)}{(3.12\times 10^5)} =\color{red}{\bf 14.0}\;\rm u$$