The air pressure on the top of the cylinder is $P_0 $. The pressure at the bottom of the pipe is $P=P_0+\rho g h$. Balancing forces on the cylinder gives
$(P-P_0)A=(\rho g h)\pi(R^2-r^2) = mg$
The total mass of the water is $750gm$, so the total volume of water in the cylinder + pipe is $750cm^3$.
Volume in pipe is $(\pi h)r^2=200cm^3$. Volume in cylinder is $750-200=550cm^3$ This is also equal to $\pi R^2H=16\pi H$ so
$H=550/16\pi \approx 11 cm$.
This agrees with the given answer of $11/32\pi$, which is in metres.
I wonder if the question-setter intended that the mass of the hole should be removed from the 3kg mass of the cylinder. However, this would not make much difference to the answer.