Answer:
(a) 2.542 cm
(b) 272.7°C
Explanation:
diameter, d = 2.540 cm
T1 = 20°C
α = 11 x 10^-6 /°C
(a) Let d' be the diameter.
T2 = 87°C
Use he formula for the areal expansion
A' = A ( 1 + βΔT)
where, β is the coefficient of areal expansion and ΔT is teh rise in temperature, A' be the area at high temperature and A be the area at low temperature.
β = 2 α = 2 x 11 x 106-6 = 22 x 10^-6 /°C
So,

D'^2 = 2.54^2 ( 1 + 22 x 10^-6 x 67)
D' = 2.542 cm
(b) Let the change in temperature is ΔT.
Use the formula for the volumetric expansion
ΔV = V x γ x ΔT
Where, γ = 3 x α = 3 x 11 x 10^-6 = 33 x 10^-6 /°C
0.9/100 = 33 x 10^-6 x ΔT
ΔT = 272.7°C