= Solution
For a twice differentiable <vector potential> $\mathbf B$, commuting partial derivatives in $\mathbf J=\nabla\times\mathbf B$ gives the necessary condition
$$
\boxed{\nabla\cdot\mathbf J=0.}
$$
The potential is not unique: for any sufficiently smooth scalar $\psi$, $\mathbf B+\nabla\psi$ has the same <curl>, since the <curl of a gradient> is zero. This is the usual <gradient> freedom of a <vector potential>. A divergence-free condition alone need not suffice globally on every domain; here only its necessity is requested.
<Stokes theorem> states that for an oriented surface with compatible boundary orientation,
$$
\int_S(\nabla\times\mathbf B)\cdot\mathbf n\,dS
=\oint_{\partial S}\mathbf B\cdot d\mathbf r.
$$
For the integrand in this problem, a convenient globally smooth <vector potential> is
$$
\mathbf B=\frac13(z^3,x^3,y^3),\qquad
\nabla\times\mathbf B=(y^2,z^2,x^2).
$$
The boundary of the upper ellipsoid is the ellipse $z=0$. The outward normal on the upper surface induces counterclockwise orientation viewed from above, so parameterize it as $\mathbf r(\theta)=(a\cos\theta,b\sin\theta,0)$, $0\leq\theta\leq2\pi$. Along this curve the first component of $\mathbf B$ is zero and $dz=0$, leaving
$$
\oint\mathbf B\cdot d\mathbf r
=\frac13\int_0^{2\pi}(a\cos\theta)^3 b\cos\theta\,d\theta
=\frac{a^3b}{3}\int_0^{2\pi}\cos^4\theta\,d\theta.
$$
Since the last angular integral is $3\pi/4$,
$$
\boxed{\int_S(y^2,z^2,x^2)\cdot d\mathbf A=\frac{\pi a^3b}{4}.}
$$
The result is independent of the vertical semiaxis $c$. As a sign check, the field is divergence-free, so closing the cap with the downward-oriented base disc makes its flux the positive integral of $x^2$ over that disc, which has the same value.
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