Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s.

Initial current, $I_{1}=5.0 \mathrm{~A}$

Final current, $l_{2}=0.0 \mathrm{~A}$

Change in current, $d I=I_{1}-I_{2}=5 \mathrm{~A}$

Time taken for the change, $t=0.1 \mathrm{~s}$

Average emf, e = 200 V

For self-inductance (L) of the coil, we have the relation for average emf as:

$e=L \frac{d i}{d t}$

$L=\frac{e}{\left(\frac{d i}{d t}\right)}$

$=\frac{200}{\frac{5}{0.1}}=4 \mathrm{H}$

Hence, the self induction of the coil is 4 H.