Aircraft pre-design gives methods to compute this, based part on physics, part on statistical data of existing aircraft. For instance the method laid out in chapter 5 of Torenbeek, following this method, we would compute required power for several cases and take the maximum.
In the aircraft design phase there is not yet data such as wing area, gross weight, fuel etc that is normally used for performance calculation, we now have the opposite problem: determine combinations of design characteristics for power plant and wing to obtain desired performance. A very detailed method is given in Torenbeek, we'll shortcut as much as possible and take the SR22 wherever we can (from the wiki and from here).
- Weight. You state an MTOW of 340 kg = 3,335 N.
- Initial estimation of aeroplane drag. Most low-speed polars can be approximated by a parabola: $$C_D = C_{D_0} + \frac{{C_L}^2}{\pi \cdot A \cdot e}$$ with A being the aspect ratio $b^2 / S$. For now, let's take statistical data listed in Torenbeek for small single aircraft with fixed gear: $ C_{D_0}$ = between 0.025 and 0.04 (take 0.035), e = 0.7. We take $ C_{D_0}$ to be on the high side because of the small size, low speed, and associated low Reynolds number with a thick friction layer. A mid range value is chosen for e. For A, let's take the value of the SR22 which is 10.1.
- Cruise: you specify 100 kts = 51.4 m/s. Horsepower $P_{CR}$ to fly at this speed and at 5,300 m altitude (ceiling of SR22): $$ P_{CR} = \frac{1}{2} \cdot \rho \cdot V^3\cdot C_D \cdot S $$ With the wing area found under 4. we get $C_L$ = 0.52 and with drag parabola from 2. $C_D$ = 0.058. Substitute $\rho$ =0.73 for 5,300m and the $P_{CR}$ = 19 kW = 26 hp at this altitude. This is net power, typical prop efficiency is 0.78 and for unboosted engines power decreases with air density. Equivalent power at sea level = (26 / 0.78) * 1.225/0.73 = 56 hp
- Stall: depending on the country there is a maximum stall speed imposed upon microlights. Your stated MTOW implies an FAA Light Sport Aircraft with a maximum stall speed of 45 knots = 23 m/s. Let's take a safety margin and take stall speed = 20 m/s. Stall speed of the SR22 = 58 kgs = 30 m/s => $$ C_{L_{max}} = \frac {2W}{\rho \cdot V^2 \cdot S} $$ = 2.0 at sea level, let's take the same $ C_{L_{max}}$ for the microlight. Substituting this value and a stall speed of 20 m/s, we then get a wing area of 6.7 m$^2$. With aspect ratio of 10.1, we get a wing span of 8.2 m
- Climb, including airworthiness requirements. Let's take the same data as the SR22, ceiling = 5,300m, rate of climb C = 6.5 m/s @ sea level. For steady state climb power $P_{climb}$: $$ \eta_p \cdot \frac{P_{climb}}{W} = C + \frac{C_D}{C_L}\cdot V $$ Minimum drag speed is minimum for $$C_L = \sqrt{3 \cdot C_{D_0} \cdot \pi \cdot A \cdot e}$$ = 1.53, equating to 23 m/s. Prop efficiency improves with airspeed and most favourable climbing speed is roughly 20% higher = 28 m/s. Typical $ \eta_p$ = 0.78 for tractor piston engine in fuselage nose. $C_L$ = 1.0 and $C_D$ = 0.08 follow from lift equation and drag polar. Resulting in:
$$ 0.78 \cdot \frac {P}{3335} = 6.5 + \frac{0.08}{1.0}\cdot 28$$
$$ P = 37.4 kW = 50 hp$$
- Take-off performance. This one is quit lengthy and involves computation of the TO field length for a given engine power - which we found under 5. so we won't be doing this exercise for now.Procedure is given in Torenbeek 5.4.5
So the power required to cruise at 100 kts at 5,300m is higher than the power required to climb: $P_{cruise}$ must be applied = 56 hp. There are many improvements that can be made on above, for more detail I refer to the book.