$\frac{5k89k2}{18}$
$18=9 \times 2$
So $5k89k2$ should be divided by $9$ as well as $2$.
As unit digit is even number $5k89k2$ is divisible by $2$.
$\textit{Divisibility rule 9:}$ Sum of digits should be multiple of $9$
$5+k+8+9+k+2 = 24 + 2k$
Now check with options
(A) $\rightarrow$ If $k = 8$ then $24 + 2*8 = 40$ not divisible by $9$.
(B) If $k = 6$ then $24 + 2*6 = 36$ divisible by $9$. Therefore $k=6$ is answer.