A submarine experiences a pressure of $$5.05 \times 10^6$$ Pa at a depth of d$$_1$$ in a sea. When it goes further to a depth of d$$_2$$, it experiences a pressure of $$8.08 \times 10^6$$ Pa. Then d$$_2$$ - d$$_1$$ is approximately (density of water = 10$$^3$$ kg/m$$^3$$ and acceleration due to gravity = 10 ms$$^{-2}$$):

To relate pressure and depth in a fluid we begin with the well-known hydrostatic formula

$$P \;=\; P_0 + \rho\,g\,h$$

where $$P$$ is the absolute pressure at depth $$h$$ below the free surface, $$P_0$$ is the pressure at the surface (atmospheric pressure), $$\rho$$ is the density of the fluid and $$g$$ is the acceleration due to gravity.

For two different depths $$d_1$$ and $$d_2$$ the corresponding pressures are

$$P_1 = P_0 + \rho g d_1$$

$$P_2 = P_0 + \rho g d_2$$

We are interested in the difference of the two depths, so we subtract the first equation from the second. The atmospheric term $$P_0$$ cancels out automatically:

$$P_2 - P_1 = \rho g (d_2 - d_1)$$

Rearranging for the depth difference gives

$$d_2 - d_1 = \frac{P_2 - P_1}{\rho g}$$

Now we substitute the numerical values given in the question. The submarine feels

$$P_1 = 5.05 \times 10^6 \text{ Pa}$$

$$P_2 = 8.08 \times 10^6 \text{ Pa}$$

and the data for water and gravity are

$$\rho = 10^3 \text{ kg m}^{-3}, \qquad g = 10 \text{ m s}^{-2}$$

First compute the pressure difference:

$$P_2 - P_1 = 8.08 \times 10^6 - 5.05 \times 10^6 = 3.03 \times 10^6 \text{ Pa}$$

Next form the product $$\rho g$$ appearing in the denominator:

$$\rho g = 10^3 \times 10 = 10^4 \text{ N m}^{-3}$$

Now divide to obtain the required depth difference:

$$d_2 - d_1 = \frac{3.03 \times 10^6}{10^4} = 3.03 \times 10^2 \text{ m} = 303 \text{ m}$$

This numerical result is very close to $$300$$ m.

Hence, the correct answer is Option C.