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An integer is said to be even integer if it can be divided into two equal groups or pairs and they are exactly divisible by 2.

In other words, even integers are multiples of 2. Symbolically, every even integer can be expressed as $2k$ for some unique integer $k$.

Consider the integers….,-4, -2, 0, 2, 4,….

Here every integer is a multiple of 2.

Also, they are exactly divisible by 2. Therefore, they are even integers.

Given that $x+3$is an even integer. Therefore, it can be expressed as multiple of 2 and therefore they are divisible by 2.

Suppose, if $x=-2$then

$\begin{array}{c}x+3=−2+3\\ =1\end{array}$

which is not divisible by 2 and hence x cannot be -1.

Suppose, if $x=-1$then

$\begin{array}{c}x+3=−1+3\\ =2\end{array}$

which is divisible by 2 and hence the value of x is 2.

Similarly, if $x=0$then

$\begin{array}{c}x+3=0+3\\ =3\end{array}$

which is not divisible by 2 and hence x cannot be 0.

If $x=2$ then

$\begin{array}{c}x+3=2+3\\ =5\end{array}$

which is not divisible by 2 and hence x cannot be 2.

Therefore, option B is correct.