$I_2$ |
$I_1$ |
$I_0$ |
|
$O_8$ |
$O_7$ |
$O_6$ |
$O_5$ |
$O_4$ |
$O_3$ |
$O_2$ |
$O_1$ |
$O_0$ |
0 |
0 |
0 |
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
|
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
1 |
|
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
|
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
|
0 |
0 |
1 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
|
0 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
0 |
1 |
1 |
1 |
|
1 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
Cube of 7 is $343$ which requires 9 bits but this can be reduced because
$I_2 = O_6$ and
$I_0 = O_0 $
7 Bits are sufficient. So the answer is $B) \ 8 \times 7$