Answer

it's the same as $7^{\frac{(2011)(2010)}{2}} = 7^{2021055}$ since $2021055 \equiv3 \pmod 4$ and by the exponential residues of $7$, we have that $7^{2021055}$ ends in $\boxed 3$

We know $7-7,7^2-9,7^3-3,7^4-1,7^5-7,...$(we write unit digit) so $7\times 7^2 \times...\times 7^{2010}=(7\times 9\times 3\times 1)^{502} \times 7 \times 9$,but $189^{502}-1$,hence the unit digit is $3$.

Note that $7^4 \equiv 1 \pmod{10}$.

it's 3 i used this : n(n+1)/2; so -> (2011)2010/2 = -> 2021055/7 = n mod 3 -> because 7^3 = 343 (last digits it's 3) so the answer it's three