Answer
$ I_0 = 0.29 A$
Work Step by Step
$ \bar P = V_{rms}I_{rms} =I_{rms}^2 R$
$ \bar P = (\frac { I_0}{\sqrt 2})^2 R = \frac{1}{2} I_0^2 R$
Ginen $\bar P = 2.0 W $ and $ R=47\Omega$
$\therefore 2.0W = \frac{1}{2} I_0^2 * 47\Omega$
$\implies I_0 = \sqrt {\frac{4.0}{47}} A = 0.29 A$
