= Solution
The relevant comparison is between <gravity> and <surface tension>. With the <mass density> difference $\Delta\rho\simeq10^3\,\mathrm{kg\,m^{-3}}$, the <Bond number> and <capillary length> are
$$
\mathrm{Bo}=\frac{\Delta\rho\,g r^2}{\gamma}
=\frac{10^3(9.81)(5\times10^{-4})^2}{0.072}\simeq0.034,
\qquad
\ell_c=\sqrt{\frac{\gamma}{\Delta\rho\,g}}\simeq2.71\,\mathrm{mm}.
$$
Thus the drop is small compared with the <capillary length>; <hydrostatic pressure> variations only weakly perturb its shape. The <Young–Laplace equation> then gives approximately constant interfacial <mean curvature>, and an axisymmetric drop on a uniform flat substrate is a <spherical cap>. The substrate selects the <contact angle>, measured through the <water>.
Write $R$ for the radius of the generating <sphere>, $s$ for the contact-circle radius, and $h$ for the apex height. Elementary <trigonometric identities> give
$$
s=R\sin\theta,\qquad h=R(1-\cos\theta)=s\tan(\theta/2).
$$
Taking the stated droplet radius to mean its radius of <curvature>, $R=r$, gives
$$
\boxed{h=0.5(1-\cos55^\circ)\,\mathrm{mm}=0.213\,\mathrm{mm}.}
$$
There is no diagram or explicit radius convention in the original PDF. If $r$ instead denotes the contact-circle radius, the answer is $h=r\tan27.5^\circ=0.260\,\mathrm{mm}$. If it denotes the radius of an equal-volume free <sphere>, use the <volume of a spherical cap>, $\pi h^2(3R-h)/3=4\pi r^3/3$, to obtain $h=r[4(1-\cos\theta)/(2+\cos\theta)]^{1/3}=0.436\,\mathrm{mm}$. The <spherical cap> approximation is justified in all three conventions; a unique numerical height requires specifying which radius is meant.
\Image[/past-exam-of-the-soft-condensed-matter-course-of-the-university-of-cambridge/2022-q1-cap.png]
{title=Geometry of the <spherical cap> and its <contact angle>}
{description=Original cross-section distinguishing the generating-sphere radius $R$, contact-circle radius $s$, and height $h$. The dashed part of the generating circle lies below the substrate.}
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