The mass defect $\Delta m$ of a nucleus $\,^{A}_{Z}\mathrm{X}$ with mass $M$ is defined as

1. The mass defect captures the fact that an assembled nucleus weighs LESS than the sum of its free constituents — the "missing" mass shows up as the nuclear binding energy. 2. syllabus defines: $\Delta m = (\text{sum of nucleon masses}) - M_{\text{nucleus}}$. 3. Sum of free nucleon masses: $Z$ protons of mass $m_p$ plus $(A-Z)$ neutrons of mass $m_n$, giving $Z m_p + (A-Z) m_n$. 4. Subtract the actual nucleus mass $M$: $\Delta m = Z m_p + (A-Z) m_n - M$. 5. By construction, $\Delta m > 0$ for any stable nucleus. Option A has the sign reversed. Options C, D don't represent the right combination of free constituents. _Source: syllabus Physics Part 2, Ch 13, §13.4 (Mass-Energy and Nuclear Binding Energy, definition of mass defect), p. 5._