$Given,$

$ \ \ $ $b = 3 m; $

$ \ \ $ $Side\ slopes = 1 : 1; $

$ \ \ $ $d = 1 m;$

$ \ \ $ $i = 1/1600 $

$ \ \ $ $N = 0.04. $

We know that the area of flow,

$$A = \frac{1}{2} \times (3 + 5) \times 1 = 4 m^2$$

and wetted perimeter,

$$P = 3 +2 \times \sqrt{1^2 + 1^2}= 5.83\ m$$

$$\therefore\ hydraulic\ mean\ depth\ m\ =\ \frac{A}{P} = \frac{4}{5.83} = 0.686\ m$$

We know that the discharge through the channel

$$Q\ =\ Area\ \times\ Velocity\ $$

$$=\ A\ \times\ \frac{1}{N}\ \times\ m^{2/3} \times i^{1/2} $$

$$ =\ 4 \times \frac{1}{0.04} \times (0.686)^{2/3} \times\ (1/1600)^{1/2}$$

$$ =\ 1.945\ m^3/s\ Ans.\ $$