c
From Bernoulli's theorem,
${P_0} + \frac{1}{2}\rho v_1^2\rho gh = {P_0} + \frac{1}{2}\rho v_2^2 + 0$
${v_2} = \sqrt {v_1^2 + 2gh} = \sqrt {0.16 + 2 \times 10 \times 0.2} = 2.03m/s$
From equation of continuity
${A_2}{v_2} = {A_1}{v_1}$
$\pi \frac{{D_2^2}}{4} \times {v_2} = \pi \frac{{D_1^2}}{4}{v_1}$
$ \Rightarrow {D_1} = {D_2}\sqrt {\frac{{{v_1}}}{{{v_2}}}} = 3.55 \times {10^{ - 3}}m$