A

如图所示，过点$D$作作$DE$、$DF$、$DG$分别垂直于$AC$，$AB$、$BC$，垂足分别为$E$、$F$、$G$，连接$AD$


$\because \angle ACB$与$\angle ABC$的角平分线交于点$D$，
$\therefore DE=DF=DG$，
$\because S_{\triangle ABC}=S_{\triangle ABD}+S_{\triangle BCD}+S_{\triangle ACD}$
$\therefore \frac{1}{2}AC•DF+\frac{1}{2}AB•DE+\frac{1}{2}BC•DG=\frac{1}{2}AB•AC$
$\therefore \frac{1}{2}(AC+AB+BC)•DG=6$，
$\therefore 6DG=6$，
$\therefore DG=1$，
$\therefore$点$D$到$BC$的距离为$1$，
故选：$A$.