${27}^{40}=3^{120}$

$3^{119}={(4-1)}^{119}={}^{119}C_0{4}^{119}-{}^{119}C_1{4}^{118}+{}^{119}C_2{4}^{117}-{}^{119}C_3{4}^{116}+\mathrm{......}+(-1)$

$\therefore 3^{119}=4k-1$

$\therefore 3^{120}=12k-3=12(k-1)+9$

$\therefore$the required remainder is 9