At a certain position, earth's magnetic field of $39 \mu T$ is horizontal and directed due north. Suppose the net field is zero exactly $8.0 cm$ from the axis of a long, straight wire . Then the wire 

 The resultant  magnetic field at that point becomes zero, when wire produces equal and opposite field. So the wire must be horizontal and carry current from west to east .

$$\begin{gathered} \frac{{\mu }_0}{2\pi }\frac{i}{r}= B_H \\ \therefore i = \frac{B_{H}r}{({\mu }_0/2\pi )}=\frac{(39\times {10}^{-6})\times 0.08}{2\times {10}^{-7}} \\ i = 15.6A from west to east \end{gathered}$$