$\qquad$



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【Analysis】Since $\left(101010_{2}=(42)_{10}\right.$ is divisible by 7, every 6 bits are exactly divisible by 7.
$2014 \div 6=335 \cdots 4$, so ( $\underbrace{201010 \cdots 10}_{2014 \text { years }}$ when divided by 7 has the same remainder as $\left(1010_{2}\right.$ when divided by 7.
And $(1010)_{2}=10_{10}$, the remainder is 3