Calculate the invert level at the end of a trench given its length, grade, and starting invert level.

The grade is $1.65\%$, and the trench length is $4.85\, \text{m}$.

$$\text{Fall} = \frac{1.65}{100} \times 4.85\, \text{m}$$ $$\text{Fall} = 0.0165 \times 4.85\, \text{m}$$ $$\text{Fall} = 0.079025\, \text{m}$$

The starting Invert Level (I.L.) at Connection Point "A" is $12.049\, \text{m}$.

$$\text{I.L. at end} = \text{Starting I.L.} - \text{Fall}$$ $$\text{I.L. at end} = 12.049\, \text{m} - 0.079025\, \text{m}$$ $$\text{I.L. at end} = 11.969975\, \text{m}$$

Rounding to three decimal places, the B.T./N.B.T. (Invert Level at the end of the trench) is $\boxed{\text{11.970 m}}$.