Volume of the air bubble is given as $\mathrm{V}_{1}=1.0 \mathrm{~cm}^{3}$

$=1.0 \times 10^{-6} \mathrm{~m}^{3}$

Air bubble rises to height given as $d=40 \mathrm{~m}$

Temperature at a depth of 40m is given as $\mathrm{~T}_{1}=12^{0} \mathrm{C}=285 \mathrm{~K}$

Temperature at the surface of the lake is given as $T_{2}=35^{0} \mathrm{C}=308 \mathrm{~K}$

The pressure on the surface of the lake will be,

$P_{2}=1 \mathrm{~atm}=1 \times 1.013 \times 10^{5} \mathrm{~Pa}$

The pressure at the depth of $40 \mathrm{~m}$ will be then,

$\mathrm{P}_{1}=1 \mathrm{~atm}+\mathrm{d} \rho \mathrm{g}$

where,

$\rho=$ density of water having value $10^{3} \mathrm{~kg} / \mathrm{m}^{3}$

$\mathrm{g}=$ acceleration due to gravity having value $9.8 \mathrm{~m} / \mathrm{s}^{2}$

Hence,

$P_{1}=1.013 \times 10^{5}+40 \times 10^{3} \times 9.8$

$\mathrm{O}_{1} \mathrm{~V}_{1} / \mathrm{T}_{1}=\mathrm{P}_{2} \mathrm{~V}_{2} / \mathrm{T}_{2}$

$=493300 \times 1 \times 10^{-6} \times 308 /\left(285 \times 1.013 \times 10^{5}\right)$

$=5.263 \times 10^{-6} \mathrm{~m}^{3} \text { or } 5.263 \mathrm{~cm}^{3}$

As a result, when the air bubble reaches the surface. its volume becomes $5.263 \mathrm{~cm}^{3}$.