Answer: \(4-\sqrt3\text{ m}\).

We start with the main method. Use exact algebra with surds and rationalise only at the final simplification stage.

The volume of a cylinder is

\(V=\pi r^2h.\)

Here

\(r=2+\sqrt3,\quad V=\pi(16+9\sqrt3).\)

So

\(\pi(2+\sqrt3)^2h=\pi(16+9\sqrt3).\)

Cancel \(\pi\):

\((2+\sqrt3)^2h=16+9\sqrt3.\)

Thus

\(h=\frac{16+9\sqrt3}{(2+\sqrt3)^2}.\)

Since

\((2+\sqrt3)^2=7+4\sqrt3,\)

we have

\(h=\frac{16+9\sqrt3}{7+4\sqrt3}.\)

Rationalise the denominator:

\(h=\frac{(16+9\sqrt3)(7-4\sqrt3)}{(7+4\sqrt3)(7-4\sqrt3)}.\)

The denominator is \(49-48=1\), and the numerator is

\(112-64\sqrt3+63\sqrt3-108=4-\sqrt3.\)

Therefore the height is

\(4-\sqrt3\text{ m}.\)

This completes the solution and gives the required result.