Step 1: Convert the given kinematic viscosity.
\[
1\ \text{stoke}=10^{-4}\ \mathrm{m^2/s}
\]
\[
\nu=0.4\times10^{-4}=4\times10^{-5}\ \mathrm{m^2/s}
\]
Diameter,
\[
D=10\ \text{cm}=0.1\ \text{m}
\]
Step 2: Use Reynolds number.
\[
Re=\frac{VD}{\nu}
\]
For \(V=0.8\ \mathrm{m/s}\),
\[
Re=\frac{0.8\times0.1}{4\times10^{-5}}
=2000
\]
Since laminar flow exists only for
\[
Re\lt 2000,
\]
the flow may not remain essentially laminar at this velocity.
Hence,
\[
\boxed{0.8\ \mathrm{m/s}}
\]
Therefore,
\[
\boxed{(A)}
\]
is the correct answer.