Given vectors are $2 \hat{i}-\hat{j}+2 \hat{k}$ and $a \hat{i}+4 \hat{j}+b \hat{k}$.
Take,
$\begin{aligned}
& \cos \theta=\frac{(2 \hat{i}-\hat{j}+2 \hat{k})(a \hat{i}+4 \hat{j}+b \hat{k})}{\sqrt{2^2+(-1)^2+2^2} \sqrt{a^2+4^2+b^2}} \\
& \frac{2}{3}=\frac{2 a-4+2 b}{\sqrt{4+1+4} \sqrt{a^2+b^2+16}} \\
& \frac{2}{3}=\frac{2(a-2+b)}{3 \sqrt{a^2+b^2+16}} \\
& \sqrt{a^2+b^2+16}=((a+b)-2)
\end{aligned}$
Take square both sides.
$\begin{aligned}
& a^2+b^2+16=(a+b)^2+4-4(a+b) \\
& a^2+b^2+16=a^2+b^2+2 a b+4-4 a-4 b \\
& 4 a+4 b+12=2 a b \\
& 2 a+2 b+6=a b \\
& 2(a+b+3)=a b
\end{aligned}$
Therefore, required value is $a b$.