Initial temperature, $T_{1}=27.0^{\circ} C$

Diameter of the hole at $T_{1}, d_{1}=4.24 cm$

Final temperature, $T_{2}=227^{\circ} C$

Diameter of the hole at $T_{2}=d_{2}$

Co-efficient of linear expansion of copper, $\alpha cu =1.70 \times 10^{-5} K ^{-1}$

For co-efficient of superficial expansion $\beta$, and change in temperature $\Delta T$, we have the relation:

$\frac{\text { Change in area }(\Delta A)}{\text { Original area }(A)}=\beta \Delta T$

$\frac{\left(\pi \frac{d_{2}^{2}}{4}-\pi \frac{d_{1}^{2}}{4}\right)}{\left(\pi \frac{d_{1}^{2}}{4}\right)}=\frac{\Delta A}{A}$

$\therefore \frac{\Delta A}{A}=\frac{d_{2}^{2}-d_{1}^{2}}{d_{1}^{2}}$

But $\beta=2 \alpha$

$\therefore \frac{d_{2}^{2}-d_{1}^{2}}{d_{1}^{2}}=2 \alpha \Delta T$

$\frac{d_{2}^{2}}{d_{1}^{2}}-1=2 \alpha\left(T_{2}-T_{1}\right)$

$\frac{d_{2}^{2}}{(4.24)^{2}}=2 \times 1.7 \times 10^{-5}(227-27)+1$

$d_{2}^{2}=17.98 \times 1.0068=18.1$

$\therefore d_{2}=4.2544 cm$

Change in diameter $=d_{2}-d_{1}=4.2544-4.24=0.0144 cm$

Hence, the diameter increases by $1.44 \times 10^{-2}\; cm .$