The speed of water leaving hose $=\sqrt {2gh}$
where h is height to be reached by water.
Let diameter of pipe='d', then
$A \times V=$Volume of water ejected per second
$=\large\frac{1}{4}$$ \pi d^2 \times \sqrt {2gh} \large\frac{m^3}{s}$
Mass of water ejected = Volume $\times$ density
$'m'=\large\frac{1}{4}$$ \pi d^2 \times \sqrt {2gh} \times density$
KE per unit time =Power of pump
$=\large\frac{1}{2}$$ mv^2=\large\frac{1}{8}$$\pi d^2 \times (2gh)^{3/2} \times density$
substituting values for h,g,density,diameter of pipe , we get
$Power =21.5\; KW$
Hence a is the correct answer.