d
The initial velocity of the water stream is given as,
$v=\sqrt{2 g(H-h)}$
The time taken by stream to hit the surface is,
$h=\frac{1}{2} g t^{2}$
$t=\sqrt{\frac{2 h}{g}}$
The horizontal range can be calculated as,
Horizontal range $=v t$
$=\sqrt{2 g(H-h)} \times \sqrt{\frac{2 h}{g}}$
$=2 \sqrt{(H-h) h}$
$=2 \sqrt{(1-0.25)(0.25)}$
$=0.866 m$