The terminal velocity achieved by ball in a viscous fluid is
$$
v_{\mathrm{t}}=\frac{2(\rho-\sigma) \mathrm{r}^2 \mathrm{~g}}{9 \eta}
$$
where, $\rho=$ density of metal of ball,
$\sigma=$ density of viscous medium,
$\mathrm{r}=$ radius of ball and
$\eta=$ coefficient of viscosity of medium
Terminal velocity of first ball,
$$
\begin{aligned}
\mathrm{v}_{\mathrm{t}_1} & =\frac{2\left(\rho_1-\sigma\right) \mathrm{r}_1^2 \mathrm{~g}}{9 \eta} \\
& =\frac{2\left(8 \rho_2-\sigma\right) \mathrm{r}_1^2 \mathrm{~g}}{9 \eta}
\end{aligned}
$$
...(i) $\left[\because \rho_1=8 \rho_2\right]$
Similarly, for second ball
$$
\mathrm{v}_{\mathrm{t}_2}=\frac{2\left(\rho_2-\sigma\right) \mathrm{r}_2^2 \mathrm{~g}}{9 \eta}
$$

From Eq. (i) and (ii), we get
$$
\begin{aligned}
\frac{\mathrm{v}_{\mathrm{t}_1}}{\mathrm{v}_{\mathrm{t}_2}} & =\frac{2\left(8 \rho_2-\sigma\right) \mathrm{r}_1^2 \mathrm{~g}}{2\left(\rho_2-\sigma\right) \mathrm{r}_2^2 \mathrm{~g}} \times \frac{9 \eta}{9 \eta} \\
& =\left(\frac{8 \rho_2-0.1 \rho_2}{\rho_2-0.1 \rho_2}\right)\left(\frac{r_1}{r_2}\right)^2
\end{aligned}
$$
..(iii) $\left[\because \sigma=0.1 \rho_2\right]$
Here, $\mathrm{r}_1=1 \mathrm{~mm}$ and $\mathrm{r}_2=2 \mathrm{~mm}$
Substituting these values in Eq. (iii), we get
$$
\Rightarrow \quad \frac{\mathrm{v}_{\mathrm{t}_1}}{\mathrm{v}_{\mathrm{t}_2}}=\left(\frac{7.9 \rho_2}{0.9 \rho_2}\right)\left(\frac{1}{2}\right)^2=\frac{79}{36}
$$