$a_{n} = 6n^{2} + 2n+a_{n-1}\rightarrow(1)$
$a_{1} = 8,a_{0} = 0$
We can rewrite the equation $(1),$ and get
$\implies a_{n}-a_{n-1} = 6n^{2} + 2n$
$a_{1}-a_{0} = 6(1)^{2} + 2(1)$
$a_{2}-a_{1} = 6(2)^{2} + 2(2)$
$a_{3}-a_{2} = 6(3)^{2} + 2(3)$
$a_{4}-a_{3} = 6(4)^{2} + 2(4)$
$\:\:\:\:\:\:\:\:\Large\vdots$
$a_{98}-a_{97} = 6(98)^{2} + 2(98)$
$a_{99}-a_{98} = 6(99)^{2} + 2(99)$
________________________________________________________
$a_{99}-a_{0} = 6\left[1^{2} + 2^{2} + 3^{2} + \dots + 99^{2} \right] + 2\left[1+2+3+\dots + 99 \right]$
$\implies K \times 10^{4} - 0 = 6\left[\dfrac{99\times 100 \times 199}{6}\right] + 2\left[\dfrac{99\times 100}{2}\right] $
$\implies K \times 10^{4} =9900 \times 199 +9900$
$\implies K \times 10^{4} =9900 (199 +1)$
$\implies K \times 10^{4} =9900 \times 200$
$\implies K \times 10^{4} =1980000$
$\implies K \times 10^{4} =198\times 10^{4}$
$\therefore K = 198$
So, the correct answer is $198.$