A long straight cable under a floor carries a steady current of 27 A…

Step 1. The field at $15$ cm. $B = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 27\ \text{A}}{0.15\ \text{m}} = 3.6 \times 10^{-5}\ \text{T}$, that is $36\ \mu\text{T}$.

Step 2. The distance for $12\ \mu\text{T}$. We want $d$, so $d = \dfrac{\mu_0 I}{2\pi B} = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 27\ \text{A}}{12 \times 10^{-6}\ \text{T}} = 0.45$ m, that is $45$ cm.