Answer
See answer.
Work Step by Step
Use the ideal gas law. The amount of gas, n, is constant, and so is the temperature. We now know from PV=nRT that the product PV is constant.
The pressure at a depth of h meters is $P=P_o+\rho gh$, where $P_o$ is atmospheric pressure and $\rho$ is the density of water.
$$(PV)_{under}=(PV)_{surface}$$
Solve for the volume at the surface.
$$V_{surface}=V_{under}\frac{P_{under}}{P_{surface}}$$
$$ =V_{under}\frac{ P_o+\rho gh }{ P_o}$$
$$ =(5.5L)\frac{1.013\times10^5 Pa+(1000kg/m^3) (9.80m/s^2)(9m) }{1.013\times10^5 Pa }\approx 10L$$
The air takes up almost twice the volume that it did before, so quickly rising to the surface without exhaling would clearly not be advisable.