$\text { P.E. }=m g h$

$=m \times 9.8 \times 10$

$=(98 \times m) J$

Also, the volume of water pumped up in time, $t$, is given as:

$V=r \times t$

$V=10^{-1} \times t m ^3$

So, the mass of the water pumped up is given by, $h$

$m=\rho \times V$

$m=10^3 kgm ^{-3} \times 10^{-1} \times tm ^3$

$m=100 \times t kg$

Now, power can be expressed as energy consumed per unit time. So, the power required to pump up the water is,

$P=\frac{P . E}{t}$

On substituting the values in the above expression, we get,

$P=\frac{98 \times 100 \times t}{t}$

$P=9800 Js ^{-1}$

$P=9800 W$

$P=9.8 kW$