= Solution
The <solid angle> is the intercepted <surface area> on a sphere of radius $R$, divided by $R^2$. Take the cone axis as the polar axis of <spherical coordinates>, and call its half-opening <angle> $\theta$. With polar integration variable $\vartheta$,
$$
\Omega=\frac1{R^2}\int_0^{2\pi}\int_0^\theta R^2\sin\vartheta\,d\vartheta\,d\phi
=\boxed{2\pi(1-\cos\theta).}
$$
For small $\theta$ measured in radians, the <Taylor series> of the <cosine> gives
$$
\boxed{\Omega=\pi\theta^2-\frac{\pi\theta^4}{12}+O(\theta^6)\simeq\pi\theta^2.}
$$
This is the <solid angle of a circular cone>, measured in <steradians>.
The supplied angular diameter is twice the half-opening <angle>, so
$$
\theta=\frac{0.533}{2}\frac\pi{180}=0.00465130246\ldots\ \mathrm{rad}.
$$
Hence either apparent disc subtends
$$
\boxed{\Omega\simeq6.797\times10^{-5}\ \mathrm{sr},\qquad
\frac{\Omega}{4\pi}\simeq5.409\times10^{-6}.}
$$
The fraction is about $0.0005409\%$ of the entire celestial sphere. Using the angular diameter directly in $\pi\theta^2$ would give four times the correct small-angle answer.
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