Relate the cone dimensions

The semi-vertical angle is $\tan \theta = \frac{3}{4}$

For the water surface at depth $h$, radius $r$ of the water cone satisfies $\frac{r}{h} = \tan \theta = \frac{3}{4}$ so $r = \frac{3h}{4}.$

Volume of water as a function of depth

Volume of a cone: $V = \frac{1}{3}\pi r^2 h.$

Substitute $r = \frac{3h}{4}$:

Differentiate with respect to time

Given $\frac{dV}{dt} = 6 \text{ m}^3/\text{hr}.$

Differentiate: $\frac{dV}{dt} = \frac{9\pi}{16} h^2 \frac{dh}{dt}.$

At $h=4$: $6 = \frac{9\pi}{16}(4)^2 \frac{dh}{dt} = 9\pi \frac{dh}{dt}.$

Hence $\frac{dh}{dt} = \frac{6}{9\pi} = \frac{2}{3\pi}.$

Wet curved surface area of water

Curved surface area of a cone is $S = \pi r l,$ where $l$ is the slant height.

Since $r:h = 3:4$, the slant height ratio is

Differentiate surface area

$\frac{dS}{dt} = \frac{15\pi}{8} h \frac{dh}{dt}.$

At $h=4$ and $\frac{dh}{dt} = \frac{2}{3\pi}$: $\frac{dS}{dt} = \frac{15\pi}{8}\cdot 4 \cdot \frac{2}{3\pi}.$

Simplify: $\frac{dS}{dt} = \frac{15\cdot 4 \cdot 2}{8\cdot 3} = 5.$

Final answer

The rate at which the wet curved surface area is increasing is $\boxed{5} \text{ m}^2/\text{hr}.$