d
$\because A=A_{0} e^{-\frac{b t}{2 m}}\left(\text { where }, A_{0}=\text { maximum amplitude }\right)$
$ According \,to \,the\, questions,\, after\, 5\, second, $
$0.9 \mathrm{A}_{0}=\mathrm{A}_{0} \mathrm{e}^{-\frac{\mathrm{b}(5)}{2 \mathrm{m}}}$ $...(i)$
$After\, 10 \,more\, second,$
$A=A_{0} \mathrm{e}^{-\frac{\mathrm{b}(15)}{2 \mathrm{m}}}$ $...(ii)$
From eq"s $(i)$ and $(ii)$
$A=0.729 A_{0} \therefore \alpha=0.729$