The rainfall on a certain day was 6 cm.

The rainfall on a certain day $=6 \mathrm{~cm}$

$=6 \times \frac{1}{100} \mathrm{~m}(\because 1 \mathrm{~m}=100 \mathrm{~cm})$

$=0.06 \mathrm{~m}$

Area of the field $=3$ hectares

We know that 1 hectare $=10000 \mathrm{~m}^{2}$

i.e., 3 hectares $=3 \times 10000 \mathrm{~m}^{2}=30000 \mathrm{~m}^{2}$

Thus, volume of rain water that fell in the field $=($ area of the field $) \times($ height of rainfall $)=30000 \times 0.06=1800 \mathrm{~m}^{3}$

Since $1 \mathrm{~m}^{3}=1000 \mathrm{~L}$, we have :

$1800 \mathrm{~m}^{3}=1800 \times 1000 \mathrm{~L}=1800000 \mathrm{~L}=18 \times 100000 \mathrm{~L}=18 \times 10^{5} \mathrm{~L}$

$\therefore$ On that day, $18 \times 10^{5} \mathrm{~L}$ of rain water fell on the field.