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 $2.512 \times 10^{-2}\; T$