b
Initially
$Q _1= CV =(2) V$
$E _1=1 / 2 CV ^2=1 / 2(2) V ^2= V ^2$
Finally
Charge on each capacitor, $Q _2=\frac{ Q _1}{2}=\frac{2 V }{2}= V$
$E _2=2\left(\frac{1}{2} \frac{ Q _2^2}{ C }\right)=\frac{ V ^2}{2} \quad \therefore \frac{ E _2}{ E _1}=\frac{1}{2}$