Step 1: Identify what fixes the resistor's voltage.
$v_i(t)$ is an ideal voltage source with the capacitor, inductor and resistor all hanging directly off it in parallel. An ideal voltage source forces its own voltage across everything connected straight to its terminals, so the resistor always has $v_i(t)$ across it, at any frequency, resonant or not.
Step 2: Pull out the amplitude.
From $v_i(t) = 12\sin(\omega t)$, the peak voltage is $V_m = 12$ V.
Step 3: Get the rms value.
For any sine wave, $V_{rms} = V_m/\sqrt2$, so here
\[ V_{rms} = \frac{12}{\sqrt2} = 6\sqrt2 \text{ V} \]
Step 4: Apply the power formula for a resistor.
\[ P = \frac{V_{rms}^2}{R} = \frac{(6\sqrt2)^2}{1000} = \frac{72}{1000} \text{ W} \]
Step 5: Convert to milliwatts.
\[ P = 0.072\text{ W} = 72\text{ mW} \]
Step 6: Note why the $1\,\mu\text{F}$ and $2.2$ mH values were a distraction here.
They would matter if we needed the current drawn from the source, since at resonance the tank's impedance peaks and the source current is smallest. But the question only asks for power in the resistor, and that is set purely by the source amplitude and $R$.
\[ \boxed{P_{avg} = 72.00\text{ mW}} \]
