Pressure at the depth, P = 80.0 atm = 80 × 1.01 × 10${}^5$ Pa
The density of water at the surface, ${\mathrm{\rho }}_1$= 1.03 × 10${}^3$ kg m${}^{-3}$
Let ρ${}_2$ be the density of water at the depth h.
Let ΔV be the change in volume.

Bulk modulus, B=$\frac{{\mathrm{PV}}_1}{∆\mathrm{V}}$$\Rightarrow \frac{∆\mathrm{V}}{{\mathrm{V}}_1}=\frac{\mathrm{P}}{\mathrm{B}}$

Compressibility of water =$\frac{1}{\mathrm{B}}=45.8\times {10}^{-11}<mspace\; width="1em"></mspace>{\mathrm{Pa}}^{-1}$

$\therefore <mspace\; width="1em"></mspace>\frac{∆\mathrm{V}}{{\mathrm{V}}_1}=80\times 1.013\times {10}^5\times 45.8\times {10}^{-11}=3.71\times {10}^{-3}$

Change in volume is given by:

$$\begin{gathered} ∆\mathrm{V}={\mathrm{V}}_1-{\mathrm{V}}_2=\mathrm{m}\left(\frac{1}{{\mathrm{\rho }}_1}-\frac{1}{{\mathrm{\rho }}_2}\right) \\ \therefore <mspace\; width="1em"></mspace>\mathrm{Volumetric}<mspace\; width="1em"></mspace>\mathrm{strain}<mspace\; width="1em"></mspace>=\frac{∆\mathrm{V}}{{\mathrm{V}}_1}=\mathrm{m}\left(\frac{1}{{\mathrm{\rho }}_1}-\frac{1}{{\mathrm{\rho }}_2}\right)\times \frac{{\mathrm{\rho }}_1}{\mathrm{m}} \\ \therefore <mspace\; width="1em"></mspace>\frac{∆\mathrm{V}}{{\mathrm{V}}_1}=1-\frac{{\mathrm{\rho }}_1}{{\mathrm{\rho }}_2} \end{gathered}$$

$$\begin{gathered} \Rightarrow 1-\frac{{\mathrm{\rho }}_1}{{\mathrm{\rho }}_2}=3.71\times {10}^{-3} \\ \Rightarrow {\mathrm{\rho }}_2=\frac{1.03\times {10}^3}{1-3.71\times {10}^{-3}}=1.034\times {10}^3{\mathrm{Kgm}}^{-3} \end{gathered}$$

Therefore, the density of water at the given depth is $1.034\times {10}^3<mspace\; width="1em"></mspace>{\mathrm{Kgm}}^{-3}.$