$$\sqrt{10 + 4\sqrt{3 -2\sqrt{2}}} = a + \sqrt{2}b$$, then what is the value of $$\sqrt{a^{3} + b^{3} + 4a^{2}b}$$?

$$\sqrt{\ 3-2\sqrt{\ 2}}$$ is the same as $$\sqrt{\ 2}-1$$
So we can rewrite the equation as, $$\sqrt{\ 10+4\sqrt{\ \left(\sqrt{\ 2}-1\right)^2}}=\sqrt{\ 10+4\left(\sqrt{\ 2}-1\right)}$$
$$\sqrt{\ 6+4\sqrt{\ 2}}=a+\sqrt{\ 2}b$$
Squaring on both side, we get
$$6+4\sqrt{\ 2}=a^2+2b^2+2\sqrt{\ 2}ab$$
Equating the rational and irrational parts on both side, we get
$$a^2+2b^2=6$$  and
$$2ab\ =\ 4$$

Trying different values, we can reach that a=2 and b=1

Putting these values in the given formula we get, 
$$\sqrt{\ a^3+b^3+4a^2b}$$
$$\sqrt{\ 8+1+16}\ $$ 
=$$\sqrt{\ 25}=5$$

Therefore, option B is the correct answer.