Given that when vector $$\vec{A}=2 \hat{i}+3 \hat{j}+2 \hat{k}$$ is subtracted from vector $$\overrightarrow{\mathrm{B}}$$, it gives a vector equal to $$2 \hat{j}$$.

We can write this as:

$$\vec{B} - \vec{A} = 2 \hat{j}$$

Now, let's express the vector $$\overrightarrow{\mathrm{B}}$$ in terms of its components:

$$\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$$

Subtract vector $$\vec{A}$$ from vector $$\vec{B}$$:

$$(B_x \hat{i} + B_y \hat{j} + B_z \hat{k}) - (2 \hat{i}+3 \hat{j}+2 \hat{k}) = 2 \hat{j}$$

Comparing the components, we get:

$$ B_x - 2 = 0 \\ B_y - 3 = 2 \\ B_z - 2 = 0 $$

Solving these equations, we find the components of vector $$\vec{B}$$:

$$ B_x = 2 \\ B_y = 5 \\ B_z = 2 $$

Now, we can find the magnitude of vector $$\vec{B}$$:

$$ |\vec{B}| = \sqrt{B_x^2 + B_y^2 + B_z^2} = \sqrt{2^2 + 5^2 + 2^2} = \sqrt{4 + 25 + 4} = \sqrt{33} $$

Therefore, the magnitude of vector $$\overrightarrow{\mathrm{B}}$$ is $$\sqrt{33}$$