No sketch is necessary here. We know the flowrate and channel width; thus we can use Eq. 15.41a to calculate the critical depth as

y_C=\left[\frac{Q^2}{gw^2} \right]^{1/3}=\left[\frac{(50\ m^3/s)^2 }{(9.81\ m/s^2)(10\ m)^2} \right]=1.37\ m

The critical velocity is found from Eq. 15.41c as

V_C=\sqrt{gy_C}=\sqrt{(9.81\ m/s^2)(1.37\ m)}=3.67\ m/s