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 come to the following.
- Weight. You state an MTOW of 340 kg.
- 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 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.
- Evaluation of performance requirements. Normally the performance is computed from aircraft data, but in the pre-design phase we 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 take the SR22 as an example wherever we can.
- Cruise: you specify 100 kts = 51.4 m/s.
- 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 kts @ 5300 m = 30 m/s => $ C_{L_{max}} = \frac {2W}{\rho \cdot V^2 \cdot S} $ = 0.36, 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 6.5 m$^2$
- 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: $ \eta_p \cdot \frac{P}{W} = C + \frac{C_D}{C_L}\cdot V $ Let's take optimum climbing speed at about halfway stall & cruise = 30 m/s. Typical $ \eta_p$ = 0.78 for tractor piston engine in fuselage nose. $C_L$ = 0.095 and $C_D$ = 0.037 follow from lift equation and drag polar. Resulting in:
$$ 0.78 \cdot \frac {P}{340} = 6.5 + \frac{0.037}{0.095}\cdot 30$$
$$ P = 8 kW$$
Again, the above is an Order Of Magnitude estimate, but it gives the method followed. There are many improvements that can be made on above, for instance differentiate the drag polar to get optimum climb speed instead of just taking an average - but for more detail I refer to the book.