1. Current when resistors are in series

Each resistor has resistance $10\,\Omega$ and there are $n$ of them.

So the equivalent external resistance in series is $R_s = 10n$

Battery emf $E = 20\,\text{V}$ and internal resistance $r = 10\,\Omega$.

Hence total resistance in the circuit is $R_{\text{total, series}} = 10n + 10$

Therefore current observed is $I = \frac{E}{10n+10} = \frac{20}{10(n+1)} = \frac{2}{n+1}$

2. Current when resistors are in parallel

For $n$ equal resistors of $10\,\Omega$ in parallel, $R_p = \frac{10}{n}$

Including internal resistance, total resistance becomes $R_{\text{total, parallel}} = 10 + \frac{10}{n}$

So new current is $I' = \frac{20}{10+\frac{10}{n}}$

Simplify: $I' = \frac{20}{10\left(1+\frac{1}{n}\right)} = \frac{2}{1+\frac{1}{n}} = \frac{2n}{n+1}$

3. Given that current increases 20 times

So, $I' = 20I$

Substitute expressions for $I'$ and $I$: $\frac{2n}{n+1} = 20\cdot \frac{2}{n+1}$

Cancel $\frac{2}{n+1}$ from both sides: $n = 20$

4. Final answer

$\boxed{20}$