90ohm differential impedance HELP

I used the formula below to calculate the differential impedance with
trace width (w) = 1.2953
trace separation (d) = .2
trace thickness (t) = .035
dielectric thickness (h) = 1.6
relative dielectric constant (er) = 4.5
Zd = [174/(sqrt(er+1.41))] * ln[5.98h/(.8w+t)] * [1- 0.48exp(-.96d/h)]
With those values, I get the differential impedance to be exactly 90ohms.

However, the width is either 1.2953mm (too big) or 1.2953mil (too small)

I'm designing a PCB... what should I do?

Thank You

Not sure what your question is. The dimension you give (all mm.) does produce 90 ohm differential impedance.

I assume you don't like the wide 1.2953 mm. runners. If that is your issue your problem is the substrate is too thick.

the dielectric thickness (i assume) to be the thickness of the PCB. Am I wrong?

and yes the problem is the 1.2953mm wide traces... for usb data... isn't that way too thick?

how can I achieve 90 impedance with small traces?

Just read what RCinFLA said. To get lower width you need to reduce the thickness of the board. IIRC last time I was doing quite long traces for usb and they were 50mil each with 9mil separation on a 1.5mm board and they worked just fine.
Generally you either have wide traces, or a thin and floppy board, or you use four layers. Pick one.