Length of the solenoid is given as I $=60 \mathrm{~cm}$
Layers of windings is $3$
Each layer has 300 turns
Number of turns per unit length, $n=(3 \times 300) / 0.6=1500 \mathrm{~m}^{-1}$
Inside the solenoid, magnetic field is given by, $\mathrm{B}=\mu_{0} \mathrm{nl}$
$\mu_{0}=4 \pi \times 10^{-7} \mathrm{~T}$
$B=\left(4 \pi \times 10^{-7}\right) \times 1500 \times 1$
$=6 \pi \times 10^{-4} \mid$
Force due to magnetic field is given by,
$F=\mid$ ‘Bl
$l^{\prime}=$ Current in the wire $=6 \mathrm{~A}$
I $=$ length of the wire $=2 \mathrm{~cm}$
When the force due to the magnetic field within the solenoid balances the weight of the wire, the windings of the solenoid would sustain the weight of the wire.
$\left.\right|^{\prime} B \mid=m g$
$m=$ mass of the wire $=2.5 \mathrm{~g}$
$\mathrm{B}=\mathrm{mg} / \mathrm{l}^{\prime} \mid$
From the previous equation (1) we get,
$6 \pi \times 10^{-4} \mathrm{l}=\mathrm{mg} / \mathrm{l}^{\prime} \mid$
$\Rightarrow \mathrm{I}=\mathrm{mg} /\left(6 \pi \times 10^{-4}\right) \mathrm{l}^{\prime} \mid$
$I=\frac{2.5 \times 10^{-3} \times 9.8}{6 \times 0.02 \times 4 \pi \times 10^{-7} \times 1500}=108.37 \mathrm{~A}$
A current of $108.37 \mathrm{~A}$ will support the wire