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**Hint:**The vectors A and B are coinitial. Coinitial vectors are those vectors which have the same starting point. Now, the two vectors terminate at the given square. The first vector A is perpendicular to the square and the vector B is not perpendicular to the square but at point B. So, there must be some angle between the two vectors.

**Complete step by step answer:**

Given, the initial point of \[\overrightarrow{A}\And \overrightarrow{B}\]is at distance of 4m. from the center of square plane,

\[\left| A \right|=4m\]

If the side of the square is a then the length of the diagonal will be \[\sqrt{2}a\]. Also, the diagonals bisect each other at right angles, so half of the length of the diagonal is \[\dfrac{a}{\sqrt{2}}\]. Now here in this problem, the length of the side of the square is \[3\sqrt{2}\] m, so half of the diagonal length will be 3m. The length of the vector \[\overrightarrow{B}\] can be find out by using Pythagoras theorem:

Let us assume the vector B meets the point D of the square and the centre of the square is O and the initial point of starting of the two vectors is P, then

$\Rightarrow P{{D}^{2}}=A{{O}^{2}}+O{{D}^{2}} \\

\Rightarrow P{{D}^{2}}={{4}^{2}}+{{3}^{2}} \\

\Rightarrow PD=\sqrt{16+9} \\

\Rightarrow PD=5 \\$

So, the magnitude of vector B is 5 m. Now taking the ratio of the magnitude of the vectors comes out to be

\[\therefore \left| A \right|:\left| B \right|=4:5\]

**So, the correct option is B.**

**Note:**Most of the time we get confused that if the magnitude is zero then the given quantity cannot be termed as a vector but in actual it depends upon the state of the quantity. Here we just need to find the magnitude of the two vectors and take the ration. First vector A was perpendicular and the distance between the centre and the starting point of the vectors equals the magnitude of vector A.

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