See V/F= flux 
Now Case I : Voltage is constant

if we increase the frequency thn flux will increase which 
ultimately increases the hysteriss loss so in order to 
dessipate the loss we hav to increase the transformer size.

NOW Case II: Flux is constant

In this case also size of transformer will increase with 
increase in frequency bcoz voltage will increase in order 
to maintain the constant flux. Hence more insulation will 
be reqd so size ultimately increases.

X(1 KVA) 50 HZ   n    Y(1 KVA ) 500 HZ 
E = 4.44 f * FLUX * N
that is emf durectly proportional to flux which is f*B*A
B- flux density n A= area of c/s of transformer

as KVA reting of both are same n hence 
X (f*B*A) = Y (f*B*A)
as frequency of X is smaller than Y so area of c/s of X 
must be larger 
hence 50 HZ transformer is largest and so on 

kishoresahoo@ymail.com

bhai kishore see here we r talking about same KVA rating 
not same KV rating.
here you r equting the KV rating(ie by equating the induced 
emfs) 
there is a diff between kv and kva hence kindly refer again 
to your solution which acco. to me is not correct

with increase in frequency the size decreases so the 50 hz 
will have big size 
for better understanding refer high frequency transformers

while increasing freq  width of stamping decreases so 50hz 
having bigest size