Answer: \(S=\dfrac{9}{8\pi}\).

The cylinder is open at both ends, so its curved surface area is

\(S=2\pi rh.\)

Since \(h=4r\),

\(S=2\pi r(4r)=8\pi r^2.\)

Differentiate with respect to \(t\):

\(\frac{\mathrm dS}{\mathrm dt}=16\pi r\frac{\mathrm dr}{\mathrm dt}.\)

It is given that

\(\frac{\mathrm dS}{\mathrm dt}=6\frac{\mathrm dr}{\mathrm dt}.\)

So

\(16\pi r\frac{\mathrm dr}{\mathrm dt}=6\frac{\mathrm dr}{\mathrm dt}.\)

At the instant considered, \(\frac{\mathrm dr}{\mathrm dt}\neq0\), so

\(16\pi r=6.\)

Hence

\(r=\frac{3}{8\pi}.\)

Therefore

\(S=8\pi\left(\frac{3}{8\pi}\right)^2 =\frac{9}{8\pi}.\)

This completes the solution and gives the required result.