= Solution
Measure $z$ from the centre of the loop of radius $R$, toward the second loop. Between the loops the two axial fields have opposite directions. Their magnitudes are
$$
B_1(z)=\frac{\mu_0IR^2}{2(R^2+z^2)^{3/2}},
\qquad
B_2(z)=\frac{\mu_0(2I)(2R)^2}{2\{(2R)^2+(D-z)^2\}^{3/2}}.
$$
Equating them and cancelling common factors gives
$$
\{4R^2+(D-z)^2\}^{3/2}=8(R^2+z^2)^{3/2}.
$$
Taking the power $2/3$ yields $(D-z)^2=4z^2$. Since $0<z<D$, we have $D-z=2z$, and therefore
$$
\boxed{z=\frac D3}.
$$
\b[Thus the <magnetic-field cancellation between oppositely driven coaxial loops> occurs one third of the separation from the smaller loop.]
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