1. Given data

   • Resistance of coil: $R = 10\,\Omega$
   • Inductance of coil: $L = 20\,\text{mH} = 20 \times 10^{-3}\,\text{H}$
   • Battery voltage: $V = 20\,\text{V}$
   • Time in which current becomes zero after opening switch: $\Delta t = 100\,\mu s = 100 \times 10^{-6}\,\text{s} = 10^{-4}\,\text{s}$

2. Current through the coil before opening the switch

   Since the coil is carrying constant current, steady state has been reached. So the inductor behaves like a short circuit and current is determined by the resistance only: $I = \frac{V}{R} = \frac{20}{10} = 2\,\text{A}$

3. Change in current after opening the switch

   The current falls from $2\,\text{A}$ to $0$ in $10^{-4}\,\text{s}$.

   So, $\Delta I = 0 - 2 = -2\,\text{A}$

4. Average induced e.m.f.

   The induced e.m.f. across an inductor is $e = -L\frac{dI}{dt}$

   Therefore, average magnitude of induced e.m.f. is $|e_{\text{avg}}| = L\left|\frac{\Delta I}{\Delta t}\right|$

   Substituting values: $|e_{\text{avg}}| = 20 \times 10^{-3} \times \frac{2}{10^{-4}}$

   $= 20 \times 10^{-3} \times 2 \times 10^{4}$

   $= 40 \times 10^{1} = 400\,\text{V}$

5. Final answer

   $\boxed{400}$

The stored correct answer is also $400$, so it matches.