Let m_p be the mass of a proton, m_n be the mass of a neutron, M₁ be the mass of a ${ }_{10}^{20}\mathrm{Ne}$, nucleus and M₂ be the mass of a ${ }_{20}^{40}\mathrm{Ca}$ nucleus. Then 

Due to mass defect (which is finally responsible for the binding energy of the nucleus), mass of a nucleus is always less then the sum of masses of it's constituent particles ${ }_{10}^{20}\mathrm{Ne}$ is made up of 10 protons plus 10 neutrons. Therefore, mass of ${ }_{10}^{20}\mathrm{Ne}$ nucleus ${\mathrm{M}}_1<10\left({\mathrm{m}}_{\mathrm{p}}+{\mathrm{m}}_{\mathrm{n}}\right)$
Also heavier the nucleus, more is he mass defect thus 20$(\left.{\mathrm{m}}_{\mathrm{n}}+{\mathrm{m}}_{\mathrm{p}}\right)-{\mathrm{M}}_2>10\left({\mathrm{m}}_{\mathrm{p}}+{\mathrm{m}}_{\mathrm{n}}\right)-{\mathrm{M}}_1$  or $10\left({\mathrm{m}}_{\mathrm{p}}+{\mathrm{m}}_{\mathrm{n}}\right)>{\mathrm{M}}_2-{\mathrm{M}}_1$
$\begin{array}{l}\Rightarrow {\mathrm{M}}_2<{\mathrm{M}}_1+10\left({\mathrm{m}}_{\mathrm{p}}+{\mathrm{m}}_{\mathrm{n}}\right)\Rightarrow {\mathrm{M}}_2<{\mathrm{M}}_1+{\mathrm{M}}_1\\ \Rightarrow {\mathrm{M}}_2<2{\mathrm{M}}_1.\end{array}$