2022 AMC 10B Problem 8

A block of ten consecutive integers contains exactly two multiples of $7$ precisely when its first multiple of $7$ is in one of the first three positions. Thus that multiple must have units digit $1,$ $2,$ or $3.$ The multiples of $7$ with those units digits are, respectively, $$\begin{gathered}21+70j,\\42+70j,\\63+70j.\end{gathered}$$ For each expression, $j=0,1,\ldots,13$ gives a value at most $1000,$ so each class contributes $14$ blocks. Therefore, the total is $3\cdot14=42.$ Thus, the answer is B .