Calculate the atmospheric pressure at sea level.

The pressure ($P$) due to a fluid column is given by the formula $P = \rho g h$, where $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity, and $h$ is the height of the fluid column.

Given:
Barometric height at sea level ($h_{sea}$) = $76 \text{ cm} = 0.76 \text{ m}$
Density of mercury ($\rho_{Hg}$) = $13600 \text{ kg/m}^3$
Acceleration due to gravity ($g$) = $10 \text{ N/kg} = 10 \text{ m/s}^2$

Pressure at sea level ($P_{sea}$):

$$P_{sea} = \rho_{Hg} g h_{sea}$$ $$P_{sea} = 13600 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 0.76 \text{ m}$$ $$P_{sea} = 103360 \text{ Pa}$$

Barometric height at the highland ($h_{high}$) = $74 \text{ cm} = 0.74 \text{ m}$

Pressure at the highland ($P_{high}$):

$$P_{high} = \rho_{Hg} g h_{high}$$ $$P_{high} = 13600 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 0.74 \text{ m}$$ $$P_{high} = 100640 \text{ Pa}$$

The pressure difference ($\Delta P$) is the difference between the pressure at sea level and the pressure at the highland.

$$\Delta P = P_{sea} - P_{high}$$ $$\Delta P = 103360 \text{ Pa} - 100640 \text{ Pa}$$ $$\Delta P = 2720 \text{ Pa}$$

The pressure difference is caused by the column of air between sea level and the highland. This pressure difference can also be expressed as $\Delta P = \rho_{air} g H$, where $\rho_{air}$ is the density of air and $H$ is the altitude.

Given:
Density of air ($\rho_{air}$) = $1.25 \text{ kg/m}^3$

Rearrange the formula to solve for $H$:

$$H = \frac{\Delta P}{\rho_{air} g}$$ $$H = \frac{2720 \text{ Pa}}{1.25 \text{ kg/m}^3 \times 10 \text{ m/s}^2}$$ $$H = \frac{2720}{12.5} \text{ m}$$ $$H = 217.6 \text{ m}$$

The altitude of the point is:

$$\textbf{217.6 m}$$