The volume of the air bubble changes due to the change in pressure as it rises from the bottom of the lake to the surface. We can use Boyle's Law to calculate the change in volume, which states that the product of pressure and volume is constant for a given mass of confined gas held at a constant temperature:

$P_1V_1 = P_2V_2$

where $P_1$ and $V_1$ are the pressure and volume at the bottom of the lake and $P_2$ and $V_2$ are the pressure and volume at the surface of the lake.

At the bottom of the lake, the pressure is the atmospheric pressure plus the pressure due to the water column above the bubble:

$P_1 = P_{\text{atm}} + \rho gh$

where $\rho$ is the density of water, $g$ is the acceleration due to gravity, and $h$ is the height of the water column. Substituting the given values, we get:

$P_1 = 1 \times 10^{5} \text{ Pa} + 1000 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 40 \text{ m} = 5 \times 10^{5} \text{ Pa}$

At the surface of the lake, the pressure is the atmospheric pressure:

$P_2 = P_{\text{atm}} = 1 \times 10^{5} \text{ Pa}$

The initial volume of the bubble is:

$V_1 = 1 \text{ cm}^3 = 1 \times 10^{-6} \text{ m}^3$

Substituting these values into Boyle's Law and solving for $V_2$, we get:

$V_2 = \frac{P_1V_1}{P_2} = \frac{5 \times 10^{5} \text{ Pa} \times 1 \times 10^{-6} \text{ m}^3}{1 \times 10^{5} \text{ Pa}} = 5 \times 10^{-6} \text{ m}^3 = 5 \text{ cm}^3$

So, the volume of the air bubble when it reaches the surface is 5 cm³.