Answer
$$Ca^{2+}\gt Mg^{2+}\gt Be^{2+}$$
For ions carrying the same charge, ionic radii increase as we down a column in the periodic table. And in the periodic table, as we move down the column 2A, the order of the elements is $Be, Mg, Ca$.
Work Step by Step
$$Ca^{2+}\gt Mg^{2+}\gt Be^{2+}$$
As shown in the textbook, for ions carrying the same charge, ionic radii increase as we move down a column in the periodic table, since the addition of new principal quantum numbers of the outermost occupied orbital of an ion greatly increases ionic radii.
Looking at periodic table, we find that $Ca$, $Mg$ and $Be$ all belong to the same column 2A. The order of the elements from top to bottom is $Be, Mg, Ca$.
As shown above, as we move down a column in the periodic table, ionic radii increase, that explains why the radii of $Ca^{2+}$ is the largest, then the radii of $Mg^{2+}$, and smallest is the radii of $Be^{2+}$.
