Given that,
Number of turns N = 250
Length of the solenoid L = 0.2 m
Magnetic field B = 3.5 x 10-5 N/A m
The solenoid magnetic field formula is B = ╬╝oIN /L
I = BL/(╬╝oN
= (3.5 x 10-5 x 0.2)/(1.26├ù10-7 x 250)
= (0.7 x 10-5/(315 x 10-7)
= 222 mA
Therefore, the current flowing through the solenoid is 222 mA.