Magnetic field strength, B is given as $100 \times {10^{-4}}T$
Number of turns per unit length, N is given as 1000 turns/m
Current carrying capacity of the coil is given as 15 A
Permeability of free space, μ0 is given as $4\pi \times {10^{-7}}Tm{A^{-1}}$
Magnetic field is given by:
B = μ0 NI/l
⇒ NI/l = B/μ0
$ = (100 \times {10^{-4}})/(4\pi \times {10^{-7}})$
NI/l = 7961
We can now explore a possible combination. Let I = 10 A and l = 0.5 m be the current and length of the solenoid, respectively.
So we get
(N x 10)/0.5 = 7961
N = 398 turns ≈ 400 turns
The length is around 50 cm, the number of spins is about 400, and the current is about 10 A. These specifics are not unusual. Limits can be adjusted to some extent.