Since 1986 = 62*32 + 2, we have
[tex]27^{1986}\equiv\left(27^{62}\right)^{32}\cdot27^2\equiv27^2\pmod{2^6}[/tex]
by Euler's theorem, since [tex]\varphi(64)=32[/tex]. Then
[tex]27^{1986}\equiv27^2\equiv729\equiv25\pmod{64}[/tex]
which tells us the 7th digit from the right is 1.
Then
[tex]25\equiv25\pmod{2^5}[/tex]
so the next digit must be 0;
[tex]25\equiv9\pmod{2^4}[/tex]
so the next digit is 1;
[tex]9\equiv1\pmod{2^3}[/tex]
so the next digit is 1;
[tex]1\equiv1\pmod{2^2}[/tex]
so the next digit is 0;
[tex]1\equiv1\pmod2[/tex]
so the next digit is 0; and
[tex]1\equiv0\pmod1[/tex]
so the last digit is 1.
So the last 7 digits of this number in binary are 1011001.
