Given, diameter of tap, $D=1.25 \mathrm{~cm}$
$$
=1.25 \times 10^{-2} \mathrm{~m}
$$
Density of water, $\rho=10^3 \mathrm{kgm}^{-3}$
Coefficient of viscosity, $\eta=10^{-3} \mathrm{~Pa}-\mathrm{s}$
Volume of water flowing out per second
$$
\begin{aligned}
Q & =3 L / \mathrm{min} \\
& =\frac{3 \times 10^{-3}}{60}=5 \times 10^{-5} \mathrm{~m}^3 \mathrm{~s}^{-1}
\end{aligned}
$$
As, $Q=v A=v \times \frac{\pi D^2}{4} \Rightarrow v=\frac{4 Q}{D^2 \pi}$
Reynold's number is given by
$$
\begin{aligned}
R_e=\frac{\rho v D}{n} & =\frac{4 \rho Q}{\pi D \eta}=\frac{4 \times 10^3 \times 5 \times 10^{-5}}{3.14 \times 1.25 \times 10^{-2} \times 10^{-3}} \\
& =5095
\end{aligned}
$$
For Reynold's number, the flow of liquid is turbulent if $R_e>3000$.
Hence, the nature of flow of water is turbulent.