Convert $-511\mathrm{.}15$ base $10$ to our own variation of the $32$ bit IEEE floating point with the following specifications: sign bit will be bit $0,$ and bits $1$ through $8$ will be the $8$ bit exponent, and bits $9$ through $31$ will be the mantissa $($do round numbers$),$ which of the following would be our number in hexadecimal using this variation $($remember$,$ like the IEEE $754$ format, we do not store the $1.$ component of our binary number in exponential form$)?$

Here is the schematic of the variation:

bit $\backslash (31\backslash$ldots $\backslash ).$ bit $9$

bit $8$ bit $1$

bit $0$

$------$ mantissa $------$ exponent $-$sign$-$

C$3$FF$9333$

FF$26670$F

$87$FF$9333$

None of the above