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Write the equation of an ellipse by: eccentricity $\epsilon = \frac{ \sqrt{3} }{ 2}$ and the distance between the directrices equals $\frac{ 6 }{ \sqrt{3} }$.

I calc: $\frac{ a^{2} }{ c } =\frac{ 3 }{ \sqrt{3} }; ~\frac{ \sqrt{3} }{ 2}=\frac{ c }{ a }; ~a\sqrt{3}=2c; ~c=\frac{a\sqrt{3}}{2}; ~\frac{ a^{2} }{ \frac{a\sqrt{3}}{2} } =\frac{ 3 }{ \sqrt{3} };~a=6; ~c=3\sqrt{3}; ~a^{2}-c^{2}=b^{2}; ~b=\sqrt{36-3\sqrt{3}}$.

And, if $a=6$, $b$ can only be approximately equal. Help me find the mistake in my solution.

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    You miscomputed $a$ by a factor of 4. – 2017-01-25

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We know the distance between the directrices of an ellipse is $d=2\frac {a}{e} $ where $e =\frac {\sqrt {a^2-b^2}}{a} $ is its eccentricity.

Using these relations we get $$\frac {6}{\sqrt {3}} =\frac {2}{\sqrt {3}}(2a) \Rightarrow a=1.5$$ Now use the eccentricity formula to get $b $ and we have our ellipse!! Hope it helps.

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    I think $a/e$ is the distance of a directrix from the center. See http://mathworld.wolfram.com/ConicSectionDirectrix.html – 2017-01-25
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    And, $b=\sqrt{(1.5^{2})-(\frac{\sqrt{27}}{4})^{2}}$? – 2017-01-25
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    @divisor Yes you are correct. – 2017-01-25