Answer
$E = \frac{\rho~a}{3~\epsilon_0}$
Work Step by Step
Let $r$ be the position vector from the center of the sphere to a point in the cavity.
Let $d$ be the position vector from the center of the cavity to this point in the cavity.
Then, in terms of vectors: $~~r = a+d$
We can use the result of part (a) to find the electric field at any point in the cavity:
$E = \frac{\rho~r}{3~\epsilon_0}-\frac{\rho~d}{3~\epsilon_0}$
$E = \frac{\rho}{3~\epsilon_0}~(r-d)$
$E = \frac{\rho~a}{3~\epsilon_0}$
