Let $m_p$ be the mass of proton, $m_n$ the mass of neutron. $M_1$ the mass of ${}_{10}{}^{20}Ne$ nucleus and $M_2$ the mass of ${}_{20}{}^{40}Ca$ nucleus. Then

Due to mass defect (which is finally responsible for the binding energy of the nucleus), mass of a nucleus is always less than the sum of masses of its constituent particles.

${}_{10}{}^{20}Ne$ is made up of $10$ protons plus $10$ neutrons

Therefore mass of ${}_{10}{}^{20}Ne$ nucleus,

$M_1<10\left(m_p+m_n\right)$

Also heavier the nucleus more is the mass defect

Thus $20\left(m_n+m_p\right)-M_2>10\left(m_n+m_p\right)-M_1$

$$\begin{gathered} 10\left(m_n+m_p\right)>M_2-M_1 \\ {M}_2<M_1+10\left(m_n+m_p\right) \end{gathered}$$

Now since $M_1<10\left(m_p+m_n\right)$

$M_2<2M_1$