For IEEE 754 single precision representation answer would be
32840.000000 0x47004800
#include<stdio.h>
int main() {
float f = 3.284e4;
printf("%f 0x%02x%02x%02x%02x\n", f,
*((char*) &f+3),*((char*) &f +2) , *((char*) &f+1), *((char*) &f+0));
}
The above code is run on a little-endian machine and hence the byte reversal. So, the answer would be
0x47004800 = (0100 0111 0000 0000 0100 1000 0000 0000)2
32840 = (1000000001001000)2
So, in normalized representation (implicit 1), we have 15 positions to be shifted for exponent and IEEE using bias as 127, we get exponent field = 127 + 15 = 142 = (10001110)2
The number is positive, so sign bit is 0.
Thus we get sign bit followed by 8 exponent bits followed by 23 mantissa bits (after removing the leading 1)
0 100 0111 0 000 0000 0100 1000 0000 0000 = 0x47004800