Solution:

$N=8=2^{3}$

The computation of $8-$ point $\mathrm{DFT}$ using radix- 2 DITFFT involves three steps.

Given, $X(k)=\{3,0,3,0,3,0,3,0\}$

$X^{*}(k)=\{3,0,3,0,3,0,3,0\}$

Step 1: The four pairs of $X^{*}(k)$ in bit reversed order is the input for step 1

The phase factor for step 1 is $W_{2}^{0}=e^{0}=1$

The butterfly computations for step 1 are:

Output of step 1 is $\mathrm{V}(\mathrm{k})=\{6,0,6,0,0,0,0,0\}$

Step 2: The output of step 1 forms the input for second step.

The phase factor for step 2 are $W_{4}^{0}=e^{0}=1 \&$ $W_{4}^{1}=e^{-j 2 \pi / 4}=\cos \frac{2 \pi}{4}-j \sin \frac{2 \pi}{4}=0-j \times 1=-j$

The butterfly computations for step 2 are:

Output of step 2 is $\mathrm{F}(\mathrm{k})=\{12,0,0,0,0,0,0,0\}$

Step 3: The output of step 2 forms the input for third step.

The phase factor for step 3 are $W_{8}^{0}=e^{0}=1 ;$

$W_{8}^{1}=e^{-j 2 \pi / 8}=\cos \frac{2 \pi}{8}-j \sin \frac{2 \pi}{8}=\frac{1}{\sqrt{2}}-j \frac{1}{\sqrt{2}}$

$W_{8}^{2}=e^{-j 4 \pi / 8}=\cos \frac{4 \pi}{8}-j \sin \frac{4 \pi}{8}=0-j \times 1=-j$

$W_{8}^{3}=e^{-j 6 \pi / 8}=\cos \frac{6 \pi}{8}-j \sin \frac{6 \pi}{8}=\frac{-1}{\sqrt{2}}-j \frac{1}{\sqrt{2}} ;$

The butterfly computations for step 3 are:

Output of step 3 is $q(n)=\{12,0,0,0,12,0,0,0\}$

$\therefore x(n)=\operatorname{IDFT}[X(k)]$

$=\frac{1}{N} q^{*}(n)$

$=\frac{1}{8}\{12,0,0,0,12,0,0,0\}$

$=\frac{1}{8}\{12,0,0,0,12,0,0,0\}$

$=\{1.5,0,0,0,1.5,0,0,0\}$

Hence, the IDFT of X (k) is

$=\{1.5,0,0,0,1.5,0,0,0\}$