Answer
$N_{\mathrm{Cu}} \Delta E_{\mathrm{be}}=1.6 \times 10^{25} \mathrm{MeV}$
Work Step by Step
The number of atoms in copper is
\begin{align*}
N_{\mathrm{Cu}}&=\left(\frac{3.0 \mathrm{g}}{62.92960 \mathrm{g} / \mathrm{mol}}\right)\left(6.02 \times 10^{23} \mathrm{atoms} / \mathrm{mol}\right)\\
& \approx 2.9 \times 10^{22} \text { atoms }
\end{align*}
The binding energy in the atom is
\begin{aligned}
\Delta E_{\mathrm{be}} &=(29(1.007825 \mathrm{u})+34(1.008665 \mathrm{u})-62.92960 \mathrm{u})(931.5 \mathrm{MeV} / \mathrm{u}) \\
&=551.4 \mathrm{MeV}
\end{aligned}
Hence, we get the total energy in the penny is
$$N_{\mathrm{Cu}} \Delta E_{\mathrm{be}}=(551.4 \mathrm{MeV})\left(2.9 \times 10^{22}\right)=\boxed{1.6 \times 10^{25} \mathrm{MeV}}$$
