From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is:

Explanation: 

According to Ampere's circuital equation, if the electric field inside the loop is constant, ∮ B.dl  of the resultant magnetic fields along a closed, plane curve is equal to μ₀ times the total current crossing the area limited by the closed curve.

\(\oint \vec{B}. \vec{dl} = μ _{0}I \)      ----- (1)

for r > a we get: B.2πr = μ₀I  ⇒ B = μ₀I /2πr     ----- (2)

for r < a we get: B.2πr = μ₀Ir²/ a²  ⇒ B = \(\frac{\mu _{0}Ir}{2\pi a^{2}} \)    ----- (3)

Solution:

from equation 2 we get-  B∝ 1/r for r>a region which means it is decreasing for this region.

from equation 3 we get-  B∝ r for r<a region which means it is increasing for this region.

Hence option 4) is correct.