Answer
$6\times10^{5}\,V/m$
Work Step by Step
Potential difference $V=600\,V$
The separation between the plates $d=1.0\,mm=1.0\times10^{-3}\,m$
The magnitude of electric field $E=\frac{V}{d}$
$=\frac{600\,V}{1.0\times10^{-3}\,m}=6\times10^{5}\,V/m$
