There are five layers of windings of \(400\) turns each on the solenoid.

\(\therefore\) Total number of turns on the solenoid, \(N=5 \times 400=2000\)

Diameter of the solenoid, \(D=1.8 \,cm =0.018\, m\)

Current carried by the solenoid, \(I=8.0 \,A\)

Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,

\(B=\frac{\mu_{0} N I}{l}\)

Where,

\(\mu_{0}=\) Permeability of free space \(=4 \pi \times 10^{-7}\, T\,m \,A ^{-1}\)

\(B=\frac{4 \pi \times 10^{-7} \times 2000 \times 8}{0.8}\) \(=8 \pi \times 10^{-3}=2.512 \times 10^{-2}\, T\)

Hence, the magnitude of the magnetic field inside the solenoid near its centre is \(2.512 \times 10^{-2}\; T\)