At the triple point of water, a constant volume gas thermometer...

For a constant volume gas thermometer, the temperature $T$ is directly proportional to the pressure $P$.

$T = \dfrac{P}{P_{tr}} \times 273.16$

Given that the pressure at the triple point of water is $P_{tr} = 40 \text{ kPa}$.

The boiling point of water is 100^{}^{{}^{\circ}} \text{C}, which corresponds to $T = 373.15 \text{ K}$.

Substituting the values into the equation:

$373.15 = \dfrac{P}{40} \times 273.16$

$P = \dfrac{373.15 \times 40}{273.16}$

$P \approx 54.6 \text{ kPa}$