Fourier Transform using pairs table
Here we can use the property stating that $$\operatorname{FT}(t^nf(t))=j^n\frac{d^n}{dw^n}\operatorname{FT}(f(t))$$

In our case \$n=2, f(t)=e^{-2t}u(t)\$. So \$\operatorname{FT}(f(t))=\frac{1}{2+j\omega}\$. The second derivative of it would be \$\frac{2j^2}{(j\omega+2)^3}=\frac{-2}{(j\omega+2)^3}\$, and I will leave it to you to calculate. Multiplying it by \$j^n=-1\$,  giving us \$\frac{2}{(j\omega+2)^3}\$ as expected.