1. Current when cells are in series

The two cells are connected in series with the same polarity, so the net emf is $E_s = 1 + 2 = 3\text{ V}$

Their internal resistances also add: $r_s = 2 + 1 = 3\Omega$

External resistance is $R = 6\Omega$

So total resistance is $R_{\text{total}} = R + r_s = 6 + 3 = 9\Omega$

Hence current $I_1 = \frac{E_s}{R_{\text{total}}} = \frac{3}{9} = \frac{1}{3}\text{ A}$

2. Current when cells are in parallel

For two cells of emfs $E_1, E_2$ and internal resistances $r_1, r_2$ connected in parallel, the equivalent emf is $E_p = \frac{E_1r_2 + E_2r_1}{r_1 + r_2}$

and equivalent internal resistance is $r_p = \frac{r_1r_2}{r_1 + r_2}$

Here, $E_1=1\text{ V},\quad r_1=2\Omega,\quad E_2=2\text{ V},\quad r_2=1\Omega$

So, $E_p = \frac{(1)(1) + (2)(2)}{2+1} = \frac{1+4}{3} = \frac{5}{3}\text{ V}$

and $r_p = \frac{(2)(1)}{2+1} = \frac{2}{3}\Omega$

Now this is connected to external resistance $6\Omega$, so total resistance is $R_{\text{total}} = 6 + \frac{2}{3} = \frac{20}{3}\Omega$

Thus, $I_2 = \frac{E_p}{R_{\text{total}}} = \frac{\frac{5}{3}}{\frac{20}{3}} = \frac{5}{20} = \frac{1}{4}\text{ A}$

3. Find the ratio

$\frac{I_1}{I_2} = \frac{\frac{1}{3}}{\frac{1}{4}} = \frac{4}{3}$

Given, $\frac{I_1}{I_2} = \frac{x}{3}$

So, $\frac{x}{3} = \frac{4}{3} \implies x=4$

4. Final Answer

$\boxed{4}$

The derived answer matches the stored correct answer.