The correct option is 3.

Step-by-Step Explanation:

From the provided image, we are given the following information regarding the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$:
1. The dot product relation: $(\overrightarrow{a}-\overrightarrow{b})·(\overrightarrow{a}+\overrightarrow{b})=27$
2. The relation between their magnitudes: $|\overrightarrow{a}|=2|\overrightarrow{b}|$

Step 1: Simplify the dot product equation
Using the distributive property of the vector dot product, we can expand the left-hand side of the first equation:
$(\overrightarrow{a}-\overrightarrow{b})��(\overrightarrow{a}+\overrightarrow{b})=\overrightarrow{a}·\overrightarrow{a}+\overrightarrow{a}·\overrightarrow{b}-\overrightarrow{b}·\overrightarrow{a}-\overrightarrow{b}·\overrightarrow{b}$

Since the dot product of vectors is commutative ($\overrightarrow{a}·\overrightarrow{b}=\overrightarrow{b}·\overrightarrow{a}$), the middle terms cancel out:
$\overrightarrow{a}·\overrightarrow{b}-\overrightarrow{b}·\overrightarrow{a}=0$

Recall that the dot product of a vector with itself is the square of its magnitude:
$\overrightarrow{a}·\overrightarrow{a}=|\overrightarrow{a}{|}^2$
and
$\overrightarrow{b}·\overrightarrow{b}=|\overrightarrow{b}{|}^2$

Substituting these simplifications back into our equation gives:
$|\overrightarrow{a}{|}^2-|\overrightarrow{b}{|}^2=27$

Step 2: Substitute the magnitude relation
We are given that $|\overrightarrow{a}|=2|\overrightarrow{b}|$. Squaring both sides yields:
$|\overrightarrow{a}{|}^2=(2|\overrightarrow{b}|{)}^2=4|\overrightarrow{b}{|}^2$

Substituting $4|\overrightarrow{b}{|}^2$ in place of $|\overrightarrow{a}{|}^2$ in our simplified dot product equation gives:
$4|\overrightarrow{b}{|}^2-|\overrightarrow{b}{|}^2=27$

Step 3: Solve for the magnitude of vector b
Subtracting the terms on the left-hand side:
$3|\overrightarrow{b}{|}^2=27$

Dividing by 3:
$|\overrightarrow{b}{|}^2=9$

Since the magnitude of any vector must be a non-negative real number ($|\overrightarrow{b}|\ge 0$), taking the square root gives:
$|\overrightarrow{b}|=3$