A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each.

Length of the solenoid, l = 80 cm = 0.8 m

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 $=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1}$

$B=\frac{4 \pi \times 10^{-7} \times 2000 \times 8}{0.8}$

$=8 \pi \times 10^{-3}=2.512 \times 10^{-2} \mathrm{~T}$

Hence, the magnitude of the magnetic field inside the solenoid near its centre is $2.512 \times 10^{-2} \mathrm{~T}$.