Answer
$E = 2.00\times 10^5~V/m$
Work Step by Step
Since the capacitors are in parallel and they are connected to the battery, the potential difference across each capacitor is $600~V$
We can find the magnitude of the electric field within the dielectric of capacitor B:
$E = \frac{V}{d}$
$E = \frac{600~V}{3.00\times 10^{-3}~m}$
$E = 2.00\times 10^5~V/m$
