Water is flowing at a rate of 3.0 litre/min.

The geyser heats the water, raising the temperature from $27^{\circ} C$ to $77^{\circ} C$.

Initial temperature, $T_{1}=27^{\circ} C$

Final temperature, $T_{2}=77^{\circ} C$

$\therefore$ Rise in temperature, $\Delta T=T_{2}-T_{1}$

$=77-27=50^{\circ} C$

Heat of combustion $=4 \times 10^{4} J / g$

Specific heat of water, $c=4.2 J g ^{-1} c C ^{-1}$

Mass of flowing water, $m=3.0$ litre/min $=3000 g / min$

Total heat used, $\Delta Q=m c \Delta T$

$=3000 \times 4.2 \times 50$

$=6.3 \times 10^{5} J / min$

$\therefore$ Rate of consumption $=\frac{6.3 \times 10^{5}}{4 \times 10^{4}}=15.75 \,g / min$