Can anybody plz solve the Q in the given attachment asap.. plzAns is opn b)

Initial temperature, T₁ = 27°C
Length of the brass wire at T₁, l = 1.8 m
Final temperature, T₂ = –39°C
Diameter of the wire, d = 2.0 mm = 2 × 10^–3 m
Tension developed in the wire = F
Coefficient of linear expansion of brass, α = 2.0 × 10^–5 K^–1
Young’s modulus of brass, Y = 0.91 × 10¹¹ Pa
Young’s modulus is given by the relation:
γ = Stress / Strain  =  (F/A) / (∆L/L)
∆L = F X L / (A X Y)      ……(i)

Where,
F = Tension developed in the wire
A = Area of cross-section of the wire.
ΔL = Change in the length, given by the relation:
ΔL = αL(T₂ – T₁) … (ii)
Equating equations (i) and (ii), we get:
αL(T₂ – T₁) = FL / [ π(d/2)²X Y ]
F = α(T₂ – T₁)πY(d/2)²
F = 2 × 10^-5 × (-39-27) × 3.14 × 0.91 × 10¹¹ × (2 × 10^-3 / 2 )²
= -3.8 × 10² N
(The negative sign indicates that the tension is directed inward.)
Hence, the tension developed in the wire is 3.8 ×10² N.