Memory allocated will be word aligned in nature.
e.g. for int the address would be allocated as a multiple of
4 .... Next char would start the n+1 where n = multiple of 4.

Similarly next.... 
when we do sizeof structure we get 60 bytes...
But originally , it should be
4 + 1+ 4+ 8 + 4 *10 = 57 bytes.
The extra three bytes are from char where 3 bytes are wasted
in memory space.

please be clear ...............i could not understand ur
reply fully.... please brief it .......sorry

Sorry ..  i did not get your context....

see from the above code i want wat address will get stored in the structure pointer temp??????????????

55 BYTES OF MEMORY NEEDED.
ADDRESSS-IS BY DEFAULT

As already told by vrushali memory to a structure is always
allocated along word boundaries. So int would fetch 4 bytes
(assumed that the int in ur machine takes 4 and word is 4
bytes). Similarly char would take 1 but since the next entry
i.e. float requires 4 so char would be given 4 (3 extra )
and so on for the remaining summing upto 60 as already
indicated.

By default temp would have the base address of the structure
i.e. pointing to the first integer i.e. 'a' in our case.

when u declare a structure there is no memory allocated yet.
memory will be allocated after creating an 
instance(variable).Here, there is an instance (pointer
variable pointing to structure FAQ ). But the compiler
allocated only 4 bytes of memory for the variable temp. But
the this temp contains garbage address or  0 if it is
global. to allocated memory of size 60 byte u need to use
malloc function.

Ex: temp = malloc ( sizeof ( struct FAQ ) );

now 60 bytes of memory has been allocated from the heap. and
the starting address of this memory chunk is stored in
variable temp.