In a co-axial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero :

Ampere's Circuital Law: $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$$

For a circular Amperean loop drawn completely outside the cable (at a radius greater than the outer conductor's radius):

$$I_{\text{enclosed}} = I_{\text{central}} + I_{\text{outer}}$$

Since the two conductors carry equal currents in opposite directions:

$$I_{\text{enclosed}} = I + (-I) = 0$$

$$\implies \oint \vec{B} \cdot d\vec{l} = 0 \implies B = 0$$

Answer: Option (A): outside the cable