Answer
18.000937u.
Work Step by Step
Assume that all of the particles have negligible momentum.
The incoming kinetic energy plus the mass-energy of the proton and the oxygen should add to the mass-energy of the products. To balance the electrons, we use the mass of hydrogen for the proton.
$$KE+ m_pc^2+m_{^{18}_{8}O}c^2=m_{^{18}_{9}F}c^2+m_nc^2 $$
$$ m_{^{18}_{9}F}c^2=KE+ m_pc^2+m_{^{18}_{8}O}c^2-m_nc^2 $$
$$ m_{^{18}_{9}F} =KE/c^2+ m_p+m_{^{18}_{8}O} -m_n $$
$$ m_{^{18}_{9}F} =2.438MeV/c^2+1.007825u+17.999160u -1.008665u $$
$$ m_{^{18}_{9}F} =0.002617u+1.007825u+17.999160u -1.008665u $$
$$ m_{^{18}_{9}F} =18.000937u $$
