Given,

Initial temperature, T₁ = 27.0°C = 27 + 273.15 = 300.15 K

Diameter of the hole at T₁, d₁ = 4.24 cm

Final temperature, T₂ = 227°C = 227 + 273.15 = 500.15 K

Co-efficient of linear expansion of copper, α_Cu = 1.70 × 10^-5 K^-1

Let the diameter of the hole at T₂ be d₂.

If the co-efficient of superficial expansion is β, and the change in temperature is ΔT, then

Change in area (∆A)/Original area (A) = β∆T

∆A/A = [(πd₂²/4) - (πd₁²/4)]/(πd₁²/4)]

⇒ ∆A/A = (d₂² - d₁²)/d₁²

But β = 2α

∴ (d₂² - d₁²)/d₁² = 2α∆T

⇒ (d₂²/d₁²) -1 = 2α(T₂ – T₁)

⇒ d₂²/(4.24 cm)² – 1 = 2 × (1.7 × 10^-5 K^-1) (500.15 K – 300.15 K)

⇒ d₂²/(17.98 cm²) = 68 × 10^-4 + 1

⇒ d₂² = 17.98 cm² × 1.0068

⇒ d₂² = 18.1022 cm²

⇒ d₂ = 4.2546 cm

Change in diameter = d₂– d₁ = 4.2546 cm – 4.24 cm =0.0146 cm Hence, the diameter increases by 0.0146 cm.