Step 1: Use area formula for two sides and included angle:
\[
\text{Area} = \frac{1}{2}ab\sin B
\]
Step 2: Substitute the known values:
\[
\text{Area} = 6 + 2\sqrt{3},
a = 2(\sqrt{3} + 1),
B = 45^\circ
\]
\[
\sin 45^\circ = \frac{1}{\sqrt{2}}
\]
Step 3: Plug into the formula:
\[
\frac{1}{2} . 2(\sqrt{3} + 1) . b . \frac{1}{\sqrt{2}} = 6 + 2\sqrt{3}
\]
\[
\Rightarrow (\sqrt{3} + 1) . \frac{b}{\sqrt{2}} = 6 + 2\sqrt{3}
\]
Step 4: Solve for \( b \):
\[
b = \frac{(6 + 2\sqrt{3}) . \sqrt{2}}{\sqrt{3} + 1}
\]
Step 5: Simplify the right-hand side:
Multiply numerator and denominator by the conjugate \( \sqrt{3} - 1 \):
\[
b = \frac{(6 + 2\sqrt{3}) . \sqrt{2} . (\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{(6 + 2\sqrt{3}) . \sqrt{2} . (\sqrt{3} - 1)}{2}
\]
Now simplify the numerator:
\[
(6 + 2\sqrt{3})(\sqrt{3} - 1) = (6\sqrt{3} - 6 + 2.3 - 2\sqrt{3}) = (6\sqrt{3} - 6 + 6 - 2\sqrt{3}) = 4\sqrt{3}
\]
Then multiply by \( \sqrt{2} \) and divide by 2:
\[
b = \frac{4\sqrt{3} . \sqrt{2}}{2} = \frac{4\sqrt{6}}{2} = 2\sqrt{6}
\]
[But this conflicts with earlier result! Let's recheck simplification.]
Instead, better to rationalize without expanding:
\[
b = \frac{(6 + 2\sqrt{3}) . \sqrt{2}}{\sqrt{3} + 1}
= \frac{2\sqrt{2}(3 + \sqrt{3})}{\sqrt{3} + 1}
= 2\sqrt{2} . \frac{3 + \sqrt{3}}{\sqrt{3} + 1}
\]
Now multiply numerator and denominator by conjugate:
\[
\frac{3 + \sqrt{3}}{\sqrt{3} + 1} . \frac{\sqrt{3} - 1}{\sqrt{3} - 1}
= \frac{(3 + \sqrt{3})(\sqrt{3} - 1)}{2} = 2
\Rightarrow b = 2\sqrt{2} . 2 = \boxed{4}
\]