By what percentage will the illumination of the lamp decrease if...

Power through the resistor is given by, $P = {I}^{2}R$.

Now, ${P}_{int} = {I}^{2}intR$

and ${P}_{final} = {\left(0 .8 {I}_{int}\right)}^{2}R$

Therefore, $$ change in power =

$\dfrac{{P}_{final} - {P}_{int}}{{P}_{int}} \times 100 = \left(0 .64 - 1\right) \times 100 = - 36$