In the following I will use h, S and V respectively for height,
surface area and volume, respectively, with subscripts A,B,C,D,E
referring to the particular cones they are of.

2:3:4:5:6 = 2k:3k:4k:5k:6k for any non-zero constant k 

hA:hB:hC:hD:hE = 2k:3k:4k:5k:6k
for any positive k

Let's find the particular positive k such that 

hA=2k, hB=3k, hC=4k, hD=5k, hE=6k, 

hB = 3k
 
3.6 = 3k
1.2 = k

Then with that constant k = 1.2, we can find any of the heights.
In particular,

hC = 4k = 4(1.2) = 4.8cm

Proportions of corresponding areas are the squares of the proportions of
corresponding linear measures.

SA:SB:SC:SD:SE = 22:32:42:52:62 = 4:9:16:25:36 = 4k:9k:16k:25k:36k 
for any positive k

Let's find the positive k such that 

SA=4k, SB=9k, SC=16k, SD=25k, SE=36k, 

SB = 9k
 
45 = 9k
5 = k

Then with that constant k=5, we can find any of the surface areas.
In particular,

SD = 25k = 25(5) = 125cm2 

Proportions of corresponding volumes are the cubes of the proportions of
corresponding linear measures.

VA:VB:VC:VD:VE = 23:33:43:53:63 = 8:27:64:125:216 = 8k:27k:64k:125k:216k
for any positive k

There exists positive k such that 

VA=8k, VB=27k, VC=64k, VD=125k, VE=216k, 

But we don't need to find that value of k as we did in the other two. We
simply divide VE = 216k by VA=8k, and the k's will cancel.

 = 27 full cones of sand from cone A are needed to fill cone E.

Edwin