Bubble rises to height, $d=40 m$

Temperature at a depth of $40 m , T_{1}=12^{\circ} C =285 K$

Temperature at the surface of the lake, $T_{2}=35^{\circ} C =308 K$

The pressure on the surface of the lake:

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

The pressure at the depth of $40 m$

$P_{1}=1 atm +d\rho g$ Where $, \rho$ is the density of

water $=10^{3} kg / m ^{3} g$ is the acceleration due

to gravity $=9.8 m / s ^{2}$

$\therefore P_{1}=1.013 \times 10^{5}+40 \times 10^{3} \times 9.8=493300 Pa$

We have: $\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$

Where, $V_{2}$ is the volume of the air bubble when it reaches the surface $V_{2}=\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}$

$=\frac{(493300)\left(1.0 \times 10^{-6}\right) 308}{285 \times 1.013 \times 10^{5}}$

$=5.263 \times 10^{-6}\; m ^{3}$ or $5.263 \;cm ^{3}$

Therefore, when the air bubble reaches the surface, its volume becomes $5.263 \;cm ^{3} .$