A submarine experiences a pressure of 5.05 × 10 ^ 6 Pa at a depth...

The pressure at a depth $d$ is $p = {p}_{0} + ρgd$, here ${p}_{0}$ is the atmospheric pressure.
$∴ {p}_{1} = {p}_{0} + ρg{d}_{1}$
and ${p}_{2} = {p}_{0} + ρg{d}_{2}$
The difference in pressure is ${p}_{2} - {p}_{1} = ρg\left({d}_{2} - {d}_{1}\right)$

Putting the values in the above equation
$8.08 \times {10}^{6} - 5.05 \times {10}^{6} = {10}^{3} \times 10\left({d}_{2} - {d}_{1}\right)$
${d}_{2} - {d}_{1} = \dfrac{3.03 \times {10}^{6}}{{10}^{4}}$
$= 303m$