You need the first 24 bits of 2*10^33. The first bit is always 1, and the remaining 23 bits form the last 23 bits of the IEEE-754 single-precision floating-point number.
Now, 2*10^33 has 110 binary digits, so it is too large to calculate exactly with most tools (calculators or programming languages). We can make things a little bit easier by noting that 2*10^33 = 2*(2*5)^33 = 2^34*5^33, so the first 24 bits of our number are the same as those of 5^33, which has only 76 bits.
We can further write:
5^33 = (2^7 - 3)^11
= 2^77 - 11*3*2^70 + 55*9*2^63 - 165*27*2^56 + 330*81*2^49
- 462*243*2^42 + 462*729*2^35 - 330*2187*2^27 + ...
= 2^53 * (2^24 - 33*2^17 + 495*2^10 - 4455*2^3 + 26730/2^4
- 112266/2^11 + 336798/2^18 - 721710/2^25 + ...)
= 2^53 * (16777216 - 4325376 + 506880 - 35640 + 1670.625
- 54.817... + 1.284... - 0.0215...)
= 2^53 * 12924697.071
= 2^53 * 110001010011011100011001b
where we rounded in the last step. So the stored part of the mantissa is 10001010011011100011001. Together with the information you already have, the result is:
0 11101101 10001010011011100011001
or in hex:
76C53719
If you want it done automatically, try this website. Type 2e33 into the top text box and hit the Rounded or Not Rounded buttons to get the answer.
If you type 2000000000000000000000000000000000 into my decimal/binary converter you will get
110001010011011100011001000100100011011001001100111000110000010101101100001010000000000000000000000000000000000
Rounded to 24 significant bits -- the number of bits in a float -- this is 110001010011011100011001 (the trailing 23 bits of this are the mantissa).
