\(B=2\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{4}=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{4}=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{4}=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{4}=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{4}=\dfrac{3^{32}-1}{4}\)
=>A>B