Answer
$B(r) = \frac{\mu_0~i~(a^2-r^2)}{(2\pi~r)(a^2-b^2)}$
Work Step by Step
We can find an expression for the current enclosed in a loop of radius $r$ where $b \lt r \lt a$:
$i_{enc} = i-\frac{\pi~(r^2-b^2)}{\pi~(a^2-b^2)}~i = \frac{i~(a^2-r^2)}{(a^2-b^2)}$
We can find an expression for $B(r)$:
$\int~B\cdot ds = \mu_0~i_{enc}$
$B\cdot 2\pi~r = \frac{\mu_0~i~(a^2-r^2)}{(a^2-b^2)}$
$B(r) = \frac{\mu_0~i~(a^2-r^2)}{(2\pi~r)(a^2-b^2)}$
