What is its nuclear binding energy per nucleon? (Mass of electron = $0.0005486\ \mathrm{u}$.)

$E_{\mathrm{B}}=8.3\ \mathrm{MeV/nucleon}$

The nuclear mass of $\mathrm{{}^{27}_{13}Al}$ is equal to its atomic mass minus the mass of its electrons. Since an aluminum atom has 13 protons, it must also have 13 electrons. So,

$$\mathrm{nuclear\ mass\ of\ {}^{27}_{13}Al}=\mathrm{(atomic\ mass\ of\ {}^{27}_{13}Al)}-13\ m_{\mathrm{e}}$$

$$\mathrm{nuclear\ mass\ of\ {}^{27}_{13}Al}=26.9815\ \mathrm{u-13(0.0005486\ u)}$$

$$\mathrm{nuclear\ mass\ of\ {}^{27}_{13}Al}=26.9744\ \mathrm{u}$$

Now, the nucleus contains $13$ protons and $27-13=14$ neutrons, so the total mass of the individual nucleons is

$$M=13\ m_{\mathrm{p}}+14\ m_{\mathrm{n}}$$

$$M=13(1.00728\ \mathrm{u})+14(1.00867\ \mathrm{u})$$

$$M=27.2160\ \mathrm{u}$$

and, the mass defect of the aluminum nucleus is

$$\Delta m=M-m$$

$$\Delta m=27.2160\ \mathrm{u-26.9744\ u}$$

$$\Delta m=0.2416\ \mathrm{u}$$

Converting this mass to energy, we can see that

$$E_{\mathrm{B}}=0.2416\ \mathrm{u}\times \frac{931\ \mathrm{MeV}}{1\ \mathrm{u}}$$

$$E_{\mathrm{B}}=225\ \mathrm{MeV}$$

so the binding energy per nucleon is

$$\frac{225\ \mathrm{MeV}}{27}=8.3\ \mathrm{MeV/nucleon}$$