Answer
$I_{rms} = 2.59~A$
Work Step by Step
We can find $Z$:
$Z = \sqrt{R^2+(X_L-X_C)^2}$
$Z = \sqrt{R^2+(\omega_d~L-\frac{1}{\omega_d~C})^2}$
$Z = \sqrt{R^2+(2\pi~f_d~L-\frac{1}{2\pi~f_d~C})^2}$
$Z = \sqrt{(15.0~\Omega)^2+[(2\pi)(550~Hz)(0.025~H)-\frac{1}{(2\pi)~(550~Hz)~(4.70\times 10^{-6}~F)}]^2}$
$Z = \sqrt{(15.0~\Omega)^2+(24.83~\Omega)^2}$
$Z = 29.0~\Omega$
We can find $I_{rms}$:
$I_{rms} = \frac{\mathscr{E}_{rms}}{Z}$
$I_{rms} = \frac{75.0~V}{29.0~\Omega}$
$I_{rms} = 2.59~A$
