A diver is 18 m down in sea water of density 1030 kg/m³…

Step 1. Gauge. $1030\ \text{kg/m}^3 \times 9.8\ \text{m/s}^2 \times 18\ \text{m} = 1.82 \times 10^{5}\ \text{Pa}$.

Step 2. Absolute. $1.013 \times 10^{5}\ \text{Pa} + 1.82 \times 10^{5}\ \text{Pa} = 2.83 \times 10^{5}\ \text{Pa}$, nearly three atmospheres.

Step 3. Push. $1.817 \times 10^{5}\ \text{Pa} \times 0.015\ \text{m}^2 = 2725\ \text{N}$.