Answer: A — $aa$

The string is $x\, b\, x\, a\, x\, a\, x\, x\, a\, x\, a\, x\, b\, a\, b$ — positions $1$ through $15$.

Identify the underlying pattern.

At every odd position, we expect $x$. At every even position, we expect a letter from $\{a, b\}$. Reading the even positions $2, 4, 6, 8, 10, 12, 14$, the letters are: $b, a, a, ?, a, ?, a$ — with positions $8$ and $12$ obscured by the placeholder $x$.

The non-$x$ letters at the visible even positions plus positions $13$ and $15$ form a repeating motif. Looking at consecutive pairs of $\{a,b\}$ letters:

• Positions $2, 4$: $ba$
• Positions $13, 14$: $ba$
• Positions $14, 15$: $ab$

To make a length-$2$ pattern repeat consistently across the string, positions $7$–$8$ must hide an $aa$ block (which preserves the run $a, a, a, a$ centred around the middle), aligning with the run of $a$'s at positions $6, 9, 10, 11$.