IEEE-754 representation for float (single-precision) type is as follows$$\begin{array}{|c|c|c|}\hline\overset{0}{\text{Sign}}&\overset{1-8}{\text{Exponent}}&\overset{9-31}{\text{Mantissa}}\\\hline\end{array}$$ Thus, the exponent field is of $8$ bits and mantissa is of $23$ bits (precision is actually of $24$ bits due to an implied $1$ mandated in normalized representation; IEEE 754 also allows denormalized numbers which are close to $0$ but this is not applicable to the given question).

The exponent field also requires sign to represent fractions. IEEE 754 does this by giving a bias -- $127$ for single-precision which means we simply subtract $127$ from the represented value to get the actual value. Thus, $0$ becomes $-127$ and $255$ (maximum value representable using $8$ bits becomes $128.$

Now, coming to the given question we need to represent $-14.25$ which equals $-1110.01$ in binary.

Converting to normalized form (only one $1$ to the left of $.)$ we get

$-1110.01 = -1.11001 \times 2^{3}$

Since we omit the implied $1$ in IEEE-754 representation we get

- Mantissa bits $= 11001$
- Exponent bits $ = 11 + 0111111 = 10000010$ (Adding bias $127)$
- Sign bit $ = 1$ (since number is negative)

This will be $1\quad 10000010\quad 11001000000000000000000$

Grouping in $4$ bits to convert ot Hexadecimal we get

$1100\quad 0001\quad 0110 \quad 0100 \quad 0000 \quad 0000\quad 0000\quad 0000$

$= C1640000H$

Option A.

More Reference: http://steve.hollasch.net/cgindex/coding/ieeefloat.html