Answer
$r = 1.28~mm$
Work Step by Step
We can find an expression for the enclosed current inside a loop of radius $r$, where $r \lt 8.00~mm$:
$i_{enc} = \frac{\pi~r^2}{\pi~(0.00800~m)^2}~(25.0~A) = (390,625~A/m^2)~r^2$
We can use Ampere's law to find $r$:
$\int~B\cdot ds = \mu_0~i_{enc}$
$B\cdot 2\pi~r = (\mu_0)~(390,625~A/m^2)~r^2$
$B = \frac{(\mu_0)~(390,625~A/m^2)~r^2}{2\pi~r} = 1.00\times 10^{-4}~T$
$\frac{(\mu_0)~(390,625~A/m^2)~r}{2\pi} = 1.00\times 10^{-4}~T$
$r = \frac{(1.00\times 10^{-4}~T)(2\pi)}{(\mu_0)~(390,625~A/m^2)}$
$r = \frac{(1.00\times 10^{-4}~T)(2\pi)}{(4\pi\times 10^{-7}~H/m)~(390,625~A/m^2)}$
$r = 0.00128~m$
$r = 1.28~mm$
