Using superposition theorem, first consider only 0.6amp source alonelet I1 be the current through $400\Omega$ resistor due to it

Using current division rule:

$I_1=\dfrac{200\times0.6}{(200+400)} = 0.2A$

$\mathrm {Total \ current\ required \ through \ 400\Omega=I=0.3A}$

So let $I_2$ be the current due to  source E

$I = I_1 + I_2$

$I_2=I -I_1=0.3 - 0.2 = 0.1A$

Voltage across $400\Omega$ will be$=0.1\times 400=40V$

Using potential divider rule:

$40 =\dfrac{400\times E}{(200+400)} = 0.2A$

$E =\dfrac{40\times600}{400}=60V$