= Solution
On the sphere, both <cosmic time> and the <comoving coordinate> $r$ are constant. The positive spatial <induced metric> obtained from the <FLRW metric> is therefore $d\ell^2=a^2r^2(d\theta^2+\sin^2\theta\,d\phi^2)$. Its <area element> is the square root of its determinant:
$$
dA=a^2r^2\sin\theta\,d\theta\,d\phi.
$$
Integrating over the <spherical coordinates> gives
$$
\boxed{A=\int_0^{2\pi}\int_0^\pi a^2r^2\sin\theta\,d\theta\,d\phi=4\pi[a(t)r]^2.}
$$
Thus $ar$ is the <areal radius>. In a curved spatial slice it need not equal the proper radial distance $a\int_0^r(1-kr'^2)^{-1/2}dr'$; the angular part of the <FLRW metric>, rather than a Euclidean radial-distance assumption, establishes the <surface area>.
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