Theoretical discharge

$Q_t = A \sqrt{2gH}$

$Q_t = \frac{1}{4}\pi (0.0125^2) \sqrt{2(9.81)(5.5)}$

$Q_t = 0.001275 ~ \text{m}^3\text{/s}$

Actual discharge

$Q_a = \dfrac{\text{Volume}}{\text{time}}$

$Q_a = \dfrac{0.45}{9.5(60)}$

$Q_a = 0.000789 ~ \text{m}^3\text{/s}$

Coefficient of discharge

$C = \dfrac{Q_a}{Q_t}$

$C = \dfrac{0.000789}{0.001275}$

$C = 0.62$ ← *answer*