Given a solenoid.
Length of the solenoid,l = 60 cm = 0.60 m
Total number of turns, N = 3 x 300 = 900
Length of the wire,  l₁ = 2.0 cm = 0.02 m
mass of the wire lying inside the solenoid,m 2.5 g = 2.5 x 10^–3 kg
Current carried by the wire,I₁ = 6.0 A 

Let, current I be passed through the solenoid windings, then,
Magnetic field produced inside the solenoid due to current is

                     $\boxed{\mathrm{B} = \frac{{\mathrm{\mu }}_0\mathrm{NI}}{l}}$ 

Force acting on wire, $={\mathrm{I}}_{1 }{l}_1 \mathrm{B} = {\mathrm{I}}_{1 }{l}_{\mathit{1}}\frac{{\mathrm{\mu }}_0\mathrm{NI}}{l}$ 

The wire can be supported if the force on wire is equal to the weight of wire, i.e. 

               ${\mathrm{I}}_1{l}_1 \frac{{\mathrm{\mu }}_0\mathrm{NI}}{l}=\mathrm{mg}$

$\Rightarrow$            $$\begin{gathered} \mathrm{I} = \frac{\mathrm{mg}l}{{\mathrm{I}}_1{l}_1{\mathrm{\mu }}_0\mathrm{N}} \\ = \frac{2.5 \times {10}^{-3}\times 9.8\times 0.6}{\left(6.0\right)\times (0.02)\times (4\mathrm{\pi }\times {10}^{-7})}\times 900 \end{gathered}$$
     
                 $= 108.27 \mathrm{A}$

Given,
Current carried by wire, I = 0.40 A,
Radius of circular wire, r = 8.0 cm = $8\times {10}^{-2}\mathrm{m}$
Number of turns in the coil, $\mathrm{n} = 100$

Using the formula, we get the magnetic field at the centre of coil as

$\boxed{\mathrm{B} = \frac{{\mathrm{\mu }}_0\mathrm{nI}}{2\mathrm{r}}} = \frac{4\mathrm{\pi }\times {10}^{-7}\times 100\times 0.4}{2\times 8.0\times {10}^{-2}}\mathrm{T}$
               $= 3.1 \times {10}^{-4}\mathrm{T}.$
                        

Given,
Current carried by conductor, I = 35 A
Distance, r = 20 cm = 0.2 m

Using Ampere's circuital law we get, 

 $$\begin{gathered} \boxed{\mathrm{B} = \frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}} \\ = \frac{4\mathrm{\pi }\times {10}^{-7}\times 35}{2\mathrm{\pi }\times 0.20}\mathrm{T} \\ = 3.5 \times {10}^{-5}\mathrm{T}. \end{gathered}$$

Given, current, I= 90 A
distance ,r = 1.5 m
Thus,
the magnitude of magnetic field is given by Ampere's circuital law 

               $$\begin{gathered} \boxed{\mathrm{B} = \frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}} \\ = \frac{4\mathrm{\pi }\times {10}^{-7}\times 90}{2\mathrm{\pi }\times 1.5} \\ = 1.2 \times {10}^{-5} T \end{gathered}$$  

Applying the right-hand thumb rule, we find that the magnetic field at the observation point is directed towards south.