Let $a, b, c, d, e, f$ be the numbers in the squares shown. Then the sum of the numbers in the four lines is $1 + 2 + 3 + ... + 9 + b + n + e$ since each of the numbers in the corner squares appears in exactly one row and one column. So $45 + b + n + e = 4 \times 13 = 52$, that is $b + n + e = 7$. Hence $b, n, e$ are $1, 2, 4$ in some order.
If $b = 2$ then $a = 2$; if $b = 4$ then $a = 0$. Both cases are impossible, so $b = 1$ and $a = 3$.
This means that $n = 2$ or $n = 4$. However, if $n = 2$ then $c = 10$, so $n = 4$ and $c = 8$.
(The values of the other letters are e = 2, d = 7, f = 6.)