We know that:

7^(4a+b) = 7^b(Mod 10) and 7^(4a) = 1 (Mod 10)

3^(4a+b) = 3^b(Mod 10) and 3^(4a) = 1 (Mod 10)

1^x = 1 (Mod 10)

Thus, 11^2003 = 1 (Mod 10)

7^2004 = 1( Mod 10)

13^2005 = 3 Mod (10)

Thus,

11^2003*7^2004*13^2005

= 1*1*3(Mod 10)

= 3 (Mod 10)

Consequently, the required last digit is 3.