Four point charges $1\mu C,-2\mu C,3\mu C$  and $-2\mu C$  are arranged on the four vertices of a square of side 1 cm.  The dipole moment of this charge assembly is

$\overrightarrow{\mathrm{p}}=\sum {\mathrm{q}}_{\mathrm{i}}\overrightarrow{{\mathrm{r}}_{\mathrm{i}}}$ we can rearrange the charges to form three dipoles with charges as  (-2 and +2) ,(+1 and _1) and (-1 and +1) with first and third in opposite direction and second one is perpendicular to them.

So, it will be the resultant of ${10}^{-8}C.m$ dipoles in perpendicular directions.