Answer
277 kPa
Work Step by Step
Please see the attached image first.
Here we use the Bernoulli's equation, $P+\frac{1}{2}\rho V^{2}+\rho gy = constant$
Where $P- Pressure$, $\frac{1}{2}\rho V^{2}-Kinetic\space energy\space per\space unit \space volume$, $\rho gy- Gravitational \space potential \space energy \space per\space unit \space volume.$
Let's plug known values into this equation.
$P+0+0=P_{0}+\frac{1}{2}\rho V^{2}+0$
$P=10^{5}kg/ms^{2}+\frac{1}{2}\times10^{3}kg/m^{3}\times(18.8\space m/s)^{2}$
$P=276,720 Pa$
$P=277 kPa$
Pressure of the water = 277 kPa
