(2183)^215 find 10s place digit? how do you solve these questions please...?

Brother...i can help u with the unit digit place...!! 
see...for these kinda sums u need to know cyclicity....like in this case
2183...we consider the last digit..3..now cyclicity of 3 is 4...how??
welll......we know
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 =24(3)
Just check out....3 comes again after 4 times...so the cyclicity becomes 4....hope u get it..
NOW,
3^215 = 3^212 * 3^3 = 3^3 =2(7)
sooo....the last digit is 7!!! hope u get that!!!

I got the answer as 8...........Whats the correct ans?

Yes, the correct answer is 8 but kindly tell the method..