Answer
$\frac{m_{neucleus}}{m_{neutron}}=7.7$
Work Step by Step
Please see the attached image first.
For an elastic collision, we can write the equation below.
$V_{2}=\frac{2m_{1}}{m_{1}+m_{2}}u_{1}+\frac{(m_{1}-m_{2})}{m_{1}+m_{2}}u_{2}-(1)$
In this case, $m_{1}=mass\space of\space neutron,\space m_{2}=mass\space of\space nucleus,\space u_{2}=0$
From $(1)=\gt$
$V_{2}=\frac{2m_{1}u_{1}}{(m_{1}+m_{2})}-(2)$
Kinetic energy of Neutron $(K_{1})=\frac{1}{2}m_{1}u_{1}^{2}$
Kinetic energy of Neutron $(K_{2})=\frac{1}{2}m_{2}V_{2}^{2}$
Percentage of transferred kinetic energy $=\frac{K_{2}}{K_{1}}\times 100\%=48.4\%$
$48.4\%=\frac{\frac{1}{2}m_{2}V_{2}^{2}}{\frac{1}{2}m_{1}u_{1}^{2}}\times100-(3)$
$(2)=\gt(3)$
$48.4\%=\frac{m_{2}(\frac{2m_{1}u_{1}}{m_{1}+m_{2}})^{2}}{m_{1}u_{1}^{2}}\times100\%$
$0.484=\frac{4m_{1}^{2}m_{2}u_{1}^{2}}{(m_{1}+m_{2})^{2}}\times\frac{1}{m_{1}u_{1}^{2}}$
$4m_{1}m_{2}=0.484(m_{1}^{2}+2m_{1}m_{2}+m_{2}^{2})$
$0=m_{1}^{2}-6.26m_{1}m_{2}+m_{2}^{2}$
We can solve this quadratic equation as follows.
$m_{1}=\frac{-(-6.26m_{2})\pm \sqrt {(-6.26m_{2})^{2}-4\times1\times m_{2}^{2}}}{2}$
$m_{1}=\frac{6.26m_{2}\pm 6m_{2}}{2}$
$(+) =\gt m_{1}=6.13m_{2}\space \space \space\space \space \space \space \space (-)=\gt m_{1}=0.13m_{2}$
In here $\space m_{1}\lt m_{2}$, therefore we have to choose second result & we get,
$\frac{m_{2}}{m_{1}}=\frac{m_{neucleus}}{m_{neutron}}=7.7$
