Leaky funnel

The normal force acting on the base of the funnel due to the water will be

$N = h\rho g \pi r^2$

This normal force will balance the water's weight as well as the weight of the funnel.

$N = m_{w}g + m_{f}g$ .

As $m_{w} = [\frac{ \pi r^2 H}{3}$ - $\frac{\pi r^2 H}{4*16*3}] \rho$

$m_{w} = \frac{ 21 \pi r^2 H}{64} \rho$

Therefore $m_{f} = \frac{ 3H \rho \pi r^2}{4} - \frac{ 21H\rho \pi r^2 }{64}$ $= \frac{ 27H \rho \pi r^2}{64}$

Putting in the values we get $m_{f} = 1.22 kg$

Therefore $\lfloor{m} \rfloor = \boxed{1}$