Answer: Option A
$${\left( {264} \right)^{102}} + {\left( {264} \right)^{103}}$$ unit digit
$$\eqalign{
& {{\text{4}}^1} \to 4 \to 4 \cr
& {4^2} \to 16 \to 6 \cr
& {4^3} \to 64 \to 4 \cr} $$
Rule: When 4 has odd power, then unit digit is 4
When 4 has even power, then unit digit is 6
$$\eqalign{
& {\left( {264} \right)^{102}} + {\left( {264} \right)^{103}} \cr
& \,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \cr
& \,\,\,\,{4^{102}}\,\,\,\,\,\,\, + \,\,\,\,\,\,{4^{103}} \cr} $$
$${\text{unit digit}} = 6 + 4 = 10 \to 0$$
    (even power)   (odd power)
$$\eqalign{
& {\bf{Alternate}} \cr
& \Rightarrow {\left( {264} \right)^{102}} + {\left( {264} \right)^{103}} \cr
& \Rightarrow {\left( {264} \right)^{102}} + \left( {1 + 264} \right) \cr
& \Rightarrow {\left( {264} \right)^{102}} + 265 \cr
& {\text{Multiple of }}5\,\,{\text{and }}2 \cr
& {\text{So}},{\text{unit digit is }}0 \cr} $$