= Solution
{c}
For an observation point in the wire's plane, both $d\mathbf s$ and $\mathbf r-\mathbf s$ lie in that plane. Their cross product is normal to it, and the <Biot-Savart law> integrates only such normal vectors. Thus the <field> is perpendicular to the plane at every point off the wire.
At the centre, all four sides contribute in the same normal direction by the right-hand rule. A single side is distance $a$ from the centre, so its contribution is
$$
\frac{\mu_0I}{4\pi}\int_{-a}^a\frac{a\,ds}{(a^2+s^2)^{3/2}}
=\frac{\mu_0I\sqrt2}{4\pi a}.
$$
Multiplying by four gives
$$
\boxed{\mathbf B_{\rm centre}=\frac{\mu_0I\sqrt2}{\pi a}\,\mathbf n,}
$$
where $\mathbf n$ is the oriented normal fixed by the current.
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