= Solution
Let the column radius be $b=\sqrt{A/\pi}$, and $J=I_0/A$ be its uniform axial <current density>. Treating it as concentric shells, only the current at radii below $r$ contributes to $B(r)$: a surrounding shell has zero interior field. Thus
$$
B_\phi(r)=\frac{\mu_0(J\pi r^2)}{2\pi r}=\frac{\mu_0Jr}{2}.
$$
The magnetic force density is radially inward, $\mathbf J\times\mathbf B=-(\mu_0J^2r/2)\hat{\mathbf r}$. Radial mechanical equilibrium at a fixed height requires
$$
\frac{dp}{dr}=-\frac{\mu_0J^2r}{2},\qquad p(r)-p(b)=\frac{\mu_0J^2}{4}(b^2-r^2).
$$
Consequently
$$
\boxed{p(0)-p(b)=\frac{\mu_0J^2b^2}{4}=\frac{\mu_0I_0^2}{4\pi A}.}
$$
The pressure is higher at the centre: its outward force balances the inward magnetic pinch. Vertical hydrostatic variation is separate from this radial pressure difference.
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