Explanation:

$2$ has a repeating pattern of $4$ i.e. the unit digit values repeat after every 4 powers
$2^1 = 2$
$2^2 = 4$
$2^3 = 8$
$2^4 = 16$
$2^5 = 32$

Now,$2^{32}$ will have a unit digit as $6$

How - Divide $32$ by $4$ and we get a remainder $0$

We are looking for a number which will have a unit digit as $6$ again!

A. $2^{16}$ has unit digit as $6$ but the number is less than $2^{32}$
B. $2^{16}$ same reason as above
C. $2^{96}$ has unit digit as $6$ and is greater than $2^{32}$
D. $2^{33}$ has unit digit as $2$
E. None of the above

Hence, option C

Another Approach:

We can also use the algebraic identity $(a^3 + b^3) = (a + b)(a^2 + b^2 - ab)$
$2^{96} + 1 = (2^{32})^3 + 1^3 = (2^{32} + 1)(2^{64} + 1 - 2^{32})$

Since, $2^{96} + 1$ is divisible by $2^{32} + 1$, it will also be divisible by that number