Gọi \({\rm{H}}\) là tâm hình vuông \({\rm{ABCD}}\) nên\({\rm{SH}} \bot \left( {{\rm{ABCD}}} \right)\)

Đặt \(AB = a\,\,(a > 0)\,,\,\,{S_{ABCD}} = {a^2};\,\,BD = a\sqrt 2 \).

Tam giác SBD vuông tại \(S\) nên \(SH = \frac{{a\sqrt 2 }}{2}\).

\({{\rm{V}}_{{\rm{S}}.{\rm{ABCD}}}} = \frac{1}{3}{\rm{SH}} \cdot {{\rm{S}}_{{\rm{ABCD}}}} = \frac{{\sqrt 2 }}{6}{{\rm{a}}^3} = \frac{{\sqrt 2 }}{6} \Rightarrow {\rm{a}} = 1\)

\({{\rm{V}}_{{\rm{MACD}}}} = \frac{1}{4}\;{{\rm{V}}_{{\rm{S}}.{\rm{ABCD}}}} = \frac{{\sqrt 2 }};\,\,{\rm{HM}} = \frac{1}{2}{\rm{SB}} = \frac{1}{2}\) (vì \({\rm{SB}} = {\rm{AB}} = 1\))

\[{S_{MAC}} = \frac{1}{2}MH \cdot AC = \frac{1}{2} \cdot \frac{1}{2} \cdot \sqrt 2 = \frac{{\sqrt 2 }}{4}.\] Ta có: \(d\left( {B,\left( {MAC} \right)} \right) = d\left( {D,\left( {MAC} \right)} \right)\).

Lại có \({V_{{\rm{MACD}}}} = \frac{1}{3} \cdot d\left( {D,\left( {MAC} \right)} \right) \cdot {{\rm{S}}_{{\rm{MAC}}}} \Rightarrow d\left( {D,\left( {MAC} \right)} \right) = \frac{{3\;{{\rm{V}}_{{\rm{MACD}}}}}}{{{{\rm{S}}_{{\rm{MAC}}}}}} = \frac{1}{2}\).

Đáp án: 0,5.