For finding normalised representation we need to find unnormalised one first. So, we have:
$0.239 \times 2^{13}$ as the number. So, we find the binary equivalent of $0.239$ till $8$ digits as capacity of mantissa field is $8$ bits. We follow the following procedure:
- $0.239 \times 2 = 0.478$
- $0.478 \times 2 = 0.956$
- $0.956 \times 2 = 1.912$
- $0.912 \times 2 = 1.824$
- $0.824 \times 2 = 1.648$
- $0.648 \times 2 = 1.296$
- $0.296 \times 2 = 0.592$
- $0.592 \times 2 = 1.184$
We stop here as we have performed $8$ iterations and hence, $8$ digits of mantissa of unnormalised number is obtained. Now we have:
Mantissa of given number $=$ $0011$ $1101$
So, the number can be written as: $0.00111101\times 2^{13}$
Now we need to align the mantissa towards left to get normalised number. And in the question it is mentioned that during alignment process $0's$ will be padded in the right side as a result of mantissa alignment to left.
So, to get normalised number, we align to left $3$ times, to get new mantissa $= 11101 000$
Exponent will also decrease by $3$ hence, new exponent $= 10$
So, normalised number$ = 1.11101000 \times 2^{10}$
Actual exponent $= 10$
Given excess $64$ is used which means bias value $= 64$
So, exponent field value $=10+64 = 74$
And of course sign $bit = 0$ being a positive number.
Thus the final representation of number
$\qquad\qquad\qquad= 0\; 1001010\;11101000$
$\qquad\qquad\qquad= 0100\;1010\;1110\;1000$
$\qquad\qquad\qquad= (4AE8)_{16}$
Hence, (D) should be the correct answer.