RC airplane required engine power
Resistance is a term used in politics and electric circuits. We here talk of drag.
Your induced drag equation uses speed where the lift coefficient belongs:
$$c_{Di} = \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$
What is missing now is the zero-lift drag which is the dominant contributor to overall drag at high speed. From your top speed result this seems to be
$$c_{D0}=\frac{2\cdot P\cdot\eta_{Prop}}{v_{max}^3\cdot S_{ref}\cdot\rho} = 0.040$$
assuming a propeller efficiency of 70% and neglecting induced drag. That figure is not unreasonable. Is the landing gear retractable?
Now we can write for the total drag as a function of speed:
$$D = \frac{\rho}{2}\cdot v^2\cdot S_{ref}\cdot\left(c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}\right)$$
$$D = \frac{\rho}{2}\cdot v^2\cdot S_{ref}\cdot c_{D0} + \frac{2\cdot (m\cdot g)^2}{\rho\cdot v^2\cdot S_{ref}\cdot\pi\cdot AR\cdot\epsilon}$$
When sizing an engine the most reasonable parameter is power loading: How many Kilowatts are available per square meter of wing area and per Kilogram of airplane mass. Wingspan is a poor basis for comparison, since a larger span at the same area means a higher aspect ratio and lower demands for power. When using your figures, your power loading is 1.21 kW for 0.54 m² or 2.25 kW per m² or 0.608 kW per kg. Compare these figures to similar model aircraft to see whether you are on the right track.
Nomenclature:
$P\;\;\;\;\;\;\:$engine power
$v\;\;\;\;\;\;\;$flight speed
$\rho\;\;\;\;\;\;\;$air density
$c_{Di}\;\;\;\;\;$induced drag coefficient
$c_{D0}\;\;\;\;\:$zero lift drag coefficient
$S_{ref}\;\;\;$reference area
$AR\;\;\;\;$aspect ratio (= 6 in your case)
$\epsilon\;\;\;\;\;\;\;$span efficiency, use 0.85 for a rectangular wing
$m\;\;\;\;\;\;$airplane mass
$g\;\;\;\;\;\;\:$gravitational acceleration