Here, hash function $h(k)= kA\; \mod\; m$ and $A=5$, where collisions are resolved by quadratic probing.
Assuming, the $i^{th}$ probe position for a value $k$ is given by the function :
$h(k,i)= (kA + i^2)\; \mod \;m $ where $i = 1,2,3\ldots$
So,
$h (65,1) = (65*5 + 1^2)\; \mod\; 10 = 6$ but at this place key $6$ is present so probe next.
$h(65,2) = (65*5 + 2^2)\; \mod \;10 = 9$ yes this position is free. So, key $65$ should be mapped at $9^{th}$ position in the hash table. But since the location starts from $1,$ (as given in question description) this will be named as location $10.$
Answer: $10$