Option A: $\sqrt{2}$
$d = a_{n+1} - a_n = [\sqrt{2}(n+1)+1] - [\sqrt{2}n+1] = \sqrt{2}$
The common difference is found by subtracting the $n$th term from the $(n+1)$th term. Since $\sqrt{2}\,n$ cancels and $+1$ cancels, only $\sqrt{2}$ remains. Note that $d$ must be a constant (no $n$ in it), which immediately rules out option B.