Here, for the electrical applicance, it is given,
Power $=5\mathrm{KW}=5000\mathrm{W}$ and
potential difference (voltage)
$=200\mathrm{V}$
We know,
$\mathrm{P}=\mathrm{VI}$
or $\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{5000}{200}=25\mathrm{A}$
i.e. the appliance will draw a current of $25\mathrm{A}$ .

So, a fuse of current rating $8\mathrm{A}$ cannot be used as it cannot allow a current of 25 A to flow through it.