jmc

Without loss of generality, let the side length of the hexagon be 1 unit. Also let $u$ be the length of each of the equal sides in the removed isosceles triangles. Define points $A$, $B$, $C$, $D$, $E$, and $F$ as shown in the diagram. Triangle $CDB$ is a 30-60-90 triangle, so $CD=u/2$ and $DB=u\sqrt{3}/2$. Also, $AB=1-2u$ because $CF=1$ and $CB=AF=u$. For the resulting dodecagon to be regular, we must have $AB=2\cdot BD$. We find $$\begin{array}{rl}1-2u& =u\sqrt{3}⟹\\ 1& =2u+u\sqrt{3}⟹\\ 1& =u(2+\sqrt{3})⟹\\ \frac{1}{2+\sqrt{3}}& =u.\end{array}$$ Multiplying numerator and denominator by $2-\sqrt{3}$ to rationalize the denominator, we get $u=2-\sqrt{3}$. The area of a regular hexagon with side length $s$ is $3s^2\sqrt{3}/2$ so the area of the hexagon is $3\sqrt{3}/2$. The removed area is $6\times \frac{1}{2}(CD)(2\cdot BD)=3u^2\sqrt{3}/2$. Therefore, the fraction of area removed is $u^2$, which to the nearest tenth of a percent is $0.072=\boxed{7.2\%}$.