Answer
The lungs would expand by a factor of 1.50
Work Step by Step
We can find the gauge pressure at a depth of 5.0 meters:
$P_g = \rho~g~h = (1030~kg/m^3)(9.80~m/s^2)(5.0~m) = 50,470~Pa$
We can find the absolute pressure initially:
$P_1 = P_{atm}+P_g$
$P_1 = (1.01\times 10^5~Pa)+(50,470~Pa)$
$P_1 = 1.5147 \times 10^5~Pa$
We can find an expression for the original volume:
$P_1~V_1 = nRT$
$V_1 = \frac{nRT}{P_1}$
We can find an expression for the final volume:
$P_2~V_2 = nRT$
$V_2 = \frac{nRT}{P_2}$
We can divide $V_2$ by $V_1$:
$\frac{V_2}{V_1} = \frac{\frac{nRT}{P_2}}{\frac{nRT}{P_1}}$
$V_2 = \frac{P_1}{P_2}~V_1$
$V_2 = \left(\frac{1.5147\times 10^5~Pa}{1.01\times 10^5~Pa}\right)~V_1$
$V_2 = 1.50~V_1$
The lungs would expand by a factor of 1.50
