Question 1005907
This problem does not sound right, it looks as if the situation is not properly described, or there is a typo.


I understand that the flower vase is a hexagonal right prism, sitting on its hexagonal base;
it is filled with 512 cubic inches of water, and
the wet portion of the inner surface of the vase is 512 square centimeters.
 
The surface area of the wet hexagonal base of the vase, in square inches, can be calculated as
{{{A[B]=(3sqrt(3)/2)*x^2}}} where {{{x}}} is the length (in inches) of a side of the base.
The wet lateral surface, in square inches, can be calculated as
{{{A[L]=6xh}}} ,
where {{{x}}} is the length (in inches) of a side of the base, and
{{{h}}} is the height (in inches) of the water.
The total wet surface, in square inches, can be calculated as {{{A=(3sqrt(3)/2)*x^2+6xh}}}
The volume of the water (in cubic inches) can be calculated as
{{{V=A[B]*h=(3sqrt(3)/2)*x^2*h}}} .
 
The problem states that
{{{system((3sqrt(3)/2)*x^2+6xh=215,(3sqrt(3)/2)*x^2*h=215)}}} .
We can solve {{{(3sqrt(3)/2)*x^2*h=215}}} for {{{h}}} , and
substitute the expression found for {{{h}}} into {{{(3sqrt(3)/2)*x^2+6xh=215}}} ,
getting an equation with only {{{x}}} as the variable.
Solving that equation, we get values for {{{x}}} ,
That we can use to calculate corresponding values for {{{h}}} .
Unfortunately,
The equation to find {{{x}}} from is an ugly cubic equation,
the solutions we finds are two sets of side length and height, and
both represent ridiculous situations
(wide vases filled to a shallow depth,
or very narrow/very tall vases filled high).
 
IF the 512 square inches had been meant to be only the wet lateral surface area
(just the walls of the vase, not including the bottom),
the equation would have been simpler,
with just one solution,
but the dimensions found would be just as ridiculous.