$t_{r+1}={{}_{}{}^{21}C}_r{\left(\frac{a}{b}\right)}^{\frac{21-r}{3}}\frac{b^{\frac{r}{3}}}{a^{\frac{r}{6}}}$
$={{}_{}{}^{21}C}_r{a}^{\frac{21-r}{3}-\frac{r}{6}}{b}^{\frac{r}{3}-\frac{21-r}{3}}$

Since, power of $a$ and $b$ are the same
$\Rightarrow \frac{21-r}{3}-\frac{r}{6}=\frac{r}{3}-\frac{21-r}{3}$
$\Rightarrow \frac{2\left(21-r\right)}{3}=\frac{r}{3}+\frac{r}{6}=\frac{r}{2}$
$84-4r=3r\Rightarrow 7r=84\Rightarrow r=12$
Hence, the ${13}^{\mathrm{t}\mathrm{h}}$ term contains the same powers of $a$ and $b$ .