Answer
$
\frac{\Delta K}{K}=\frac{4 m_n m}{\left(m+m_n\right)^2}
$
Work Step by Step
The energy lost by the neutron is the same as the energy gained by the target nucleus, so
$
\Delta K=\frac{1}{2} m v_f^2=\frac{1}{2} \frac{4 m_n^2 m}{\left(m+m_n\right)^2} v_{n i}^2 .
$
The initial kinetic energy of the neutron is $K=\frac{1}{2} m_n v_{n i}^2$, so
$
\frac{\Delta K}{K}=\frac{4 m_n m}{\left(m+m_n\right)^2}
$
