= Solution
On the punctured region $r>0$, choose a positive constant $a$ and set
$$
\boxed{\psi(r)=1+\frac ar.}
$$
For a radial function in three dimensions, $\Delta\psi=r^{-2}\partial_r(r^2\partial_r\psi)$. Since $r^2\partial_r\psi=-a$, this <harmonic function> has $\Delta\psi=0$ at every $r>0$, is nonconstant and positive, and approaches one at infinity. It defines the requested <time-symmetric conformally flat vacuum initial data> on that punctured or exterior region. One often writes $a=M/2$, giving the time-symmetric spatial metric of the <Schwarzschild metric> in isotropic coordinates.
\b[A puncture or inner boundary is necessary if global regularity was intended.] A smooth <harmonic function> on all of $\mathbb R^3$ that tends to one at infinity must equal one everywhere: applying the <maximum principle for harmonic functions> to larger and larger balls bounds $|\psi-1|$ by its arbitrarily small boundary values. The nonconstant example is therefore not a smooth solution through $r=0$, and no such globally regular example exists under those stronger assumptions.
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