The number of electrons in ${Be}^{3+}$ are 1. If true enter 1, else enter 0.

Radius of ${Li}^{2+}=0.1763\stackrel{\circ }{A}{Be}^{3+}=0.1323\stackrel{\circ }{A}H{e}^{+}=0.2645\stackrel{\circ }{A}$ use equation $r_n=\frac{n^2{a}_0}{Z}$
In this case, z=$Z^{\ast }$ since each contains one electron and no screening takes place, ie., $\sigma =0$