An integer is divisible by 16 if and only if its last …… digits are divisible by 16.
$\left( a \right)3$
$\left( b \right)4$
$\left( c \right)5$
$\left( d \right)6$

Hint: In this particular question use the concept that in any number which has 4 or greater than 4 digits is divisible by 16 if and only if the last 4 digits of the number is divisible by 16, so use this concept to get the solution of this question.

Complete step-by-step answer:
The divisibility rule of 16 is given as,
Any number which has 4 or greater than 4 digits is divisible by 16 if and only if the last 4 digits of the number is divisible by 16.
For example:
4374208
As we see in above number last four digits are 4208, so divide these digits by 16 we have,
$ \Rightarrow \dfrac{{4208}}{{16}} = 263$
So according to the divisibility rule of 16, the whole number is divisible by 16.
\[ \Rightarrow \dfrac{{4374208}}{{16}} = 273388\]
Take another example
4373216
As we see in above number last four digits are 3216, so divide these digits by 16 we have,
$ \Rightarrow \dfrac{{3216}}{{16}} = 201$
So according to the divisibility rule of 16, the whole number is divisible by 16.
$ \Rightarrow \dfrac{{4373216}}{{16}} = 273326$
By observing we can say that from ${1^{st}}$ and ${2^{nd}}$ condition if the last four digits are divisible by 16 then the whole number is divisible by 16.
So the integer is divisible by 16 if and only if, if its last four digits are divisible by 16.
Hence option (B) is the correct answer.

Note: Sometimes the condition for divisibility by 16 is not remembered to there can be another approach to solve this, simply take examples of the numbers divisible by 16, maybe take multiples of 16 only and try and observe for the pattern or the divisibility of digits to know when it can be divisible by 16.