$h = 3.8 \, \text{ m}$
Neglecting head lost
$E_1 = E_3$
$\dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma} + z_1 = \dfrac{{v_3}^2}{2g} + \dfrac{p_3}{\gamma} + z_3$
$0 + 0 + 3.8 = \dfrac{8Q^2}{\pi^2g{D_3}^4} + 0 + 0$
$\dfrac{8Q^2}{\pi^2(9.81)(0.05^4)} = 3.8$
$Q = 0.01695 \, \text{ m}^3\text{/s} = 16.95 \, \text{ L/s}$ answer
Considering head lost
$HL = 0.10h = 0.10(3.8)$
$HL = 0.38 \, \text{ m}$
$E_1 - HL = E_3$
$\dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma} + z_1 - HL = \dfrac{{v_3}^2}{2g} + \dfrac{p_3}{\gamma} + z_3$
$(0 + 0 + 3.8) - 0.38 = \dfrac{8Q^2}{\pi^2g{D_3}^4} + 0 + 0$
$\dfrac{8Q^2}{\pi^2(9.81)(0.05^4)} = 3.42$
$Q = 0.01608 \, \text{ m}^3\text{/s} = 16.08 \, \text{ L/s}$ answer