Got inspiration from u @Yash_Srivastava2 bro : another way : Quantity is (3+2\sqrt{2})^3
Let a_n=(3+2\sqrt{2})^n+(3-2\sqrt{2})^n so that a_0=2, a_1=6 and a_{n+2}=6a_{n+1}-a_n
We get a_2=6^2-2=34 and a_3=6\times 34-6=198
And since (3-2\sqrt{2})^3\in(0,1), we get \left\lfloor(3+2\sqrt{2})^3\right\rfloor=a_3-1
\boxed{197}