The electronic configuration of

$B ^{+} \rightarrow \,1 s ^2 \,2 s ^2$

$Be ^{+} \rightarrow \,1 s ^2\, 2 s ^1$

$Mg ^{+} \rightarrow \,1 s ^2\, 2 s ^2\, 2 p ^6\, 3 s ^1$

$Si ^{+} \rightarrow \,1 s ^2 \,2 s ^2 \,2 p ^6 \,3 s ^2 \,3 p ^1$

So, in $B ^{+}$the electron is to be removed from completely filled $s$-subshell, which is difficult. Thus, the second ionization energy of boron is maximum.