Answer
$$0.50 \space m^3$$
Work Step by Step
Boyle's law: $p_iV_i = p_fV_f$
$1 \space atm = 101325 \space Pa$
$(100 \space cm)^3 = (1\space m)^3 \longrightarrow 10^6 \space cm^3 = 1 \space m^3$
$p_f = p_i + \rho gh$
$p_i = 1 \space atm$ (Sea level)
$$\rho gh =(1.025 \space g \space cm^{-3})(9.81 \space ms^{-2})(50 \space m) $$ $$\rho gh = 502.7625 \space \frac{g \space m^2}{cm^{3} s^2} \times \frac{10^6 \space cm^3}{1 \space m^3} \times \frac{1 \space kg}{1000 \space g}$$ $$\rho gh = 5.027625 \space 10^5 \space \frac{kg \space m}{s^2 \space m^2} = 5.027625 \space 10^5 \space Pa$$
$$p_f = 1 \space atm + 5.027625 \space 10^5 \space Pa$$ $$p_f = 101325 \space Pa + 5.027625 \space 10^5 \space Pa = 604087.5 \space Pa$$
$$(101325 \space Pa)(3.0 \space m^3) = (604087.5 \space Pa)V_f$$ $$V_f = \frac{(101325 \space Pa)(3.0 \space m^3)}{604087.5 \space Pa} = 0.50 \space m^3$$
