https://snag.gy/fIHzQr.jpg

Here's my logic

Call the Height of the cylinder, H.....note that 2m of this is already filled in the first 5 hours

Call the volume filled  in the last 4 hrs   =   

pi * [ H - 2 ]  * 3^2  =         9pi [ H - 2 ]  m^3

So....in every hour....the volume filled must just be  1/4 of this   

(9/4)pi [ H - 2 ]  m^3

Now ....call the volume of the cone filled =   pi/3 * 4 * 3^2  =  12 pi   m^3

And note that  if the sand is 6m above the vertex of the cone after 5 hrs, then 2m of the cylinder are also filled in the first 5 hours....so....the volume of the partially filled cylinder  = 

pi * 2 * 3^2  =   18pi  m^3

So....the total volume filled after 5 hours just must be  12pi + 18 pi   =  30 pi m^3

So...in one hour the volume filled  is just 1/5 of this  = 6 pi m^3

But the volume filled  every hour is the same which implies that

(9/4)pi [ H - 2 ]  m^3  =  6 pi m ^3   so

(9/4) pi [ H - 2 ]  =  6 pi         divide the  "pi's "  out

(9/4) [ H - 2 ]  =  6 pi       multiply both sides by  4/9

H - 2   =  6(4/9)

H - 2  =  24/9    =  8/ 3

H  =  8/3  + 2

H  =  8/3 + 6/3

H  =  14/3    m  =    cylinder height