Answer:
L = 0.48 H
Explanation:
let L be the inductance, Irms be the rms current, Vrms be the rms voltage and Vmax  be the maximum voltage and XL be the be the reactance of the inductor.
Vrms = Vmax/(√2)
     = (3.00)/(√2)
     = 2.121 V
then:
XL = Vrms/I Â
   = (2.121)/(2.50×10^-3)
   = 848.528 V/A
that is L = XL/(2×π×f)
       = (848.528)/(2×π×(280))
       = 0.482 H
Therefore, the inductance needed to kepp the rms current less than 2.50mA is 0.482 H.