Answer
$ \Delta E_{1}= 19.8 \mathrm{MeV}$
Work Step by Step
We add energy to produce ${ }^{4} \mathrm{He} \rightarrow p+{ }^{3} \mathrm{H},$So, the energy needed is
$$
\begin{aligned}
\Delta E_{1} &=\left(m_{3 \mu}+m_{14}-m_{4 \mathrm{He}}\right) c^{2}\\
&=(3.01605 \mathrm{u}+1.00783 \mathrm{u}-4.00260 \mathrm{u})(931.5 \mathrm{MeV} / \mathrm{u}) \\
&=\boxed{19.8 \mathrm{MeV}}
\end{aligned}
$$
