Answer
$\sqrt{2}E_{\circ}$
Work Step by Step
We know that
$P_1=\frac{1}{2}P_2$
We also know that
$E_{rms}=\frac{E_{max}}{\sqrt 2}$
As rms electric field of beam 1 has the value $E_{\circ}$
Now $E_{\circ}=(E_{rms})1$
$E_{\circ}=\frac{E_{max}}{\sqrt 2}$
$\implies (E_{rms})_2=\sqrt 2E_{\circ}$
Thus, the rms electric field of beam 2 is $\sqrt{2}E_{\circ}$
