It can be seen that the average of two consecutive primes with a gap of size $8$ between them is odd. If this average is $k^2$, then we may assume it ends with a digit $1, 5,$ or $9$. But $1$ and $9$ are discounted because adding or subtracting $8/2=4$ to or from the average will result in either of the two numbers ending with $5$, so both of them cannot be prime. Therefore the average $k^2$ itself must end with the digit $5$, and so must $k$ also.
Is there such integer $k$ that $k^2$ is the average of two consecutive primes with a gap of size $8$ between them?
The answer is no.
For any $k$, the candidate primes would be $k^2-4$ and $k^2+4$. For $k = 1$ and $k = 2$, the lower candidate is too small; for $k = 3$, the candidates are $5$ and $13$, which are both primes, but not consecutive; and for $k \geq 4$, the lower candidate factors as $k^2-4 = (k-2)(k+2)$, and is therefore not prime.