Answer
$E = 4200~N/C$
Work Step by Step
We can find the charge enclosed within a sphere of radius $3.5~cm$:
$q = \rho~V$
$q = \rho~\frac{4}{3}\pi~r^3$
$q = (3.2\times 10^{-6}~C/m^3)~(\frac{4}{3}\pi)~(0.035~m)^3$
$q = 5.747\times 10^{-10}~C$
We can find the magnitude of the electric field at $r = 3.5~cm$:
$\epsilon~\Phi = q$
$\epsilon~E~4\pi~r^2 = q$
$E = \frac{q}{\epsilon~4\pi~r^2}$
$E = \frac{5.747\times 10^{-10}~C}{(8.854\times 10^{-12}~F/m)~(4\pi)~(0.035~m)^2}$
$E = 4200~N/C$
