Since h = 1/2 r then you'll have r as:
r = 2h
note that the formula in finding the volume of a cone is defined by:
V =  \frac{1}{3}  \pi r^2 h
where r in that case is 2h. then you'll have it as:
V =  \frac{1}{3}  \pi (2h)^2 h
V =  \frac{1}{3}  \pi (4h^2) h
V =  \frac{4}{3}  \pi h^3
since you are to find the rate at which the height is increasing then you will have to differentiate the derived equivalent formula.
V =  \frac{4}{3}  \pi h^3
 \frac{dV}{dt} =  \frac{4}{3} \pi (3h^2)  \frac{dh}{dt}
 \frac{dV}{dt} = 4 \pi h^2  \frac{dh}{dt}
substituting the given you'll have:
Given: dV/dt = 1cu.m/min
                h = 6m
 \frac{dV}{dt} =  4 \pi h^2  \frac{dh}{dt}
1 = 4 \pi (6^2)  \frac{dh}{dt}
 \frac{dh}{dt} =  \frac{1}{144 \pi }
 \frac{dh}{dt} = 0.00221 \frac{m}{min}