Peltier Kits Cooling
First estimate the heat flow you have to deal with. That depends on the temperature difference, times box conductivity.
Temperature difference dt is your outside ambient (not given) less inside temperature of 15°C
Box conductivity depends on the area, thickness and construction. A cubical 90L box has an inside side of about 450 mm. Assume you make it 50 mm thick, that gives a mean side of 500 mm, so an effective area of 1.5 m2.
Assume you use the best possible commercially available material of PU insulation foam, this has a K of 22 mW.m-1, so gives a box conductivity of KA/d = 22 m * 1.5 / 50 m = 660 mW/°C
If your outside ambient is 35°C, then dt is 20°C, and the power you need to abstract from the box is then 13 watts.
Put '12706' into google, find a datasheet or online calculator, and see what power is required to shift 13 watts across a dt of 20°C, or rather more than that dt depending on the quality of both your hot side and your cold side heatsink. Their temperature drops could double dt, making the peltier require much more power. It's worth putting a lot of thought into these heatsinks, perhaps even making the hot side heatsink evaporative, to benefit from getting it nearer to, or even below, ambient. I favour using a waterblock on the hot side, and circulating the water through a bong/swamp/evaporative cooler.
Then put 'power supply' into google, and find one that will supply that.
Of course, if any of my assumptions are wrong, the numbers will be wrong. Your box may have a larger area for not being cubical. You may have thinner insulation, or insulation of not such a low K value like rockwool or polystyrene or paper, or air leaks, or cold bridging, all of which will increase the power required.
In short, you have a lot of detail to handle before you will get a power supply estimate.