= Solution
The parameters satisfy $R^2C=100\,\mu\mathrm H>L=10\,\mu\mathrm H$, and
$$
\omega_*=3.33\times10^5\,\mathrm{rad\,s^{-1}},\qquad Z_*=\frac{L}{RC}=1\,\Omega.
$$
Thus the source current is $10$ A RMS. The ideal reactive elements dissipate no average power, so all mean input power is lost in $R$:
$$
\boxed{\overline P=\frac{V_{\rm rms}^2}{Z_*}=100\,\mathrm W,\qquad P_{\rm resistor,peak}=2\overline P=200\,\mathrm W.}
$$
The source's instantaneous input power also has maximum $200$ W because its total current is in phase with its voltage, although its power maximum need not occur at the resistor's power maximum: the reactive elements exchange stored energy.
With only $R$ across the source,
$$
\boxed{\overline P_R=10\,\mathrm W,\qquad P_{R,\rm peak}=20\,\mathrm W.}
$$
The network transforms the real input load from $10\,\Omega$ to $1\,\Omega$. At $\omega_*$ the parallel-branch impedance is $1+3i\,\Omega$, giving resistor RMS voltage $10\sqrt{10}$ V and again $(10\sqrt{10})^2/10=100$ W. The extra power is supplied by the voltage source; it is not produced by the ideal capacitor or inductor.
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