If X

$X=\left\{8^{n}-7 n-1, n \in N\right\}$

$Y=\{49 n-49 ; n \in N\}$

Since $8^{n}-7 n-1=(1+7)^{n}-7 n-1$

$=\left(1+7 n+{ }^{n} C_{2} 7^{2}+{ }^{n} C_{3} 7^{3}+\ldots+7^{n}\right)-7 n-1$

$=1+7 n+{ }^{n} C_{2} 49+{ }^{n} C_{3} 7^{3}+\ldots+7^{n}-7 n-1$

$=7^{2}\left({ }^{n} C_{2}+{ }^{n} C_{3} 7+\ldots+7^{n-2}\right)$

$=49\left({ }^{n} C_{2}+{ }^{n} C_{3} 7+\ldots+7^{n-2}\right)$

i.e. $8^{n}-7 n-1$ is a multiple of 49

i. e. $X \subset Y$

Here $X=\{0,49,490,4067, \ldots\}$

$Y=\{0,49,98, \ldots \ldots\}$

$Y \not \subset X($ since $98 \notin x)$

$\therefore X \subset Y$

Hence, the correct answer is option A.