smc

Let $△ABC$ have side length $s$ and $△A^{\prime }{B}^{\prime }{C}^{\prime }$ have side length $t$. Thus the altitude of $△ABC$ is $\frac{s\sqrt{3}}{2}$. Now observe that this altitude is made up of three parts: the distance from $BC$ to $B^{\prime }{C}^{\prime }$, plus the altitude of $△A^{\prime }{B}^{\prime }{C}^{\prime }$, plus a top part which is equal to the length of the diagonal line from the bottom-left corner of $△ABC$ to the bottom left corner of $△A^{\prime }{B}^{\prime }{C}^{\prime }$ (as an isosceles trapezium is formed with parallel sides $AB$ and $A^{\prime }{B}^{\prime }$, and legs $A{A}^{\prime }$ and $B{B}^{\prime }$). We drop a perpendicular from $B^{\prime }$ to $BC$, which meets $BC$ at $D$. $△BD{B}^{\prime }$ has angles $30^{\circ }$, $60^{\circ }$, and $90^{\circ }$, and the vertical side is that distance from $BC$ to $B^{\prime }{C}^{\prime }$, which is given as $\frac{1}{6}\times \frac{s\sqrt{3}}{2}=\frac{s\sqrt{3}}{12}$, so that by the length relationships in a 30-60-90 triangle,, the length of the diagonal line is $\frac{s\sqrt{3}}{12}\times 2=\frac{s\sqrt{3}}{6}.$ Thus using the "altitude in three parts" idea, we get $\frac{s\sqrt{3}}{2}=\frac{s\sqrt{3}}{12}+\frac{t\sqrt{3}}{2}+\frac{s\sqrt{3}}{6}⟹\frac{s\sqrt{3}}{4}=\frac{t\sqrt{3}}{2}⟹t=\frac{1}{2}s.$ Thus the sides of $△A^{\prime }{B}^{\prime }{C}^{\prime }$ are half as long as $△ABC$, so the area ratio is $(\frac{1}{2}{)}^2=\frac{1}{4}$, which is $\boxed{\frac{1}{4}}$.