Answer
$I_{rms}=1.92mA$
Work Step by Step
We know that
$Z=\sqrt{R^2+(\frac{1}{\omega C})^2}$
We plug in the known values to obtain:
$Z=\sqrt{(10.0)^2+[\frac{1}{2\pi (105)(0.250)}]^2}$
$Z=11.7\Omega$
Now we can find the rms current as
$I_{rms}=\frac{rms}{Z}$
We plug in the known values to obtain:
$I_{rms}=\frac{22.5}{11.7}=1.92mA$