Given equation is, $x^3+bx^2+cx-190 = 0$; or $x^3+bx^2+cx = 190$
Converting this equation into decimal, $x^3+bx^2+cx = r^2+9r$ -------------(1), where $r$ is the original base.
Now all the roots, ie; $5$,$8$,$9$ should satisfy equation (1).
Putting $x=5\Rightarrow$ $5^3+5^2.b+5c = r^2+9r$ -------------- (2)
Putting $x=8\Rightarrow$ $8^3+8^2.b+8c = r^2+9r$ -------------- (3)
Putting $x=9\Rightarrow$ $9^3+9^2.b+9c = r^2+9r$ -------------- (4)
$\frac{(2)}{(3)}$ gives $\frac{125+25b+5c}{512+64b+8c}=1$; or $125+25b+5c = 512+64b+8c$
Solving; $-39b-3c=387$ or $13b+c=-129$ -------------------- (5)
$\frac{(3)}{(4)}$ gives $\frac{512+64b+8c}{729+81b+9c}=1$; or $512+64b+8c = 729+81b+9c$
ie; $-17b-c=217$ --------------- (6)
$(5) + (6)$ gives $-4b = 88;$ So $b=-22.$
Substituting $b$ in $(5)$, we get $c = -129 + (22*13) = 157$.
Putting the values of $b$ and $c$ in $(2)$,
$5^3+5^2.(-22)+5*157 = r^2+9r$
ie; $r^2+9r-125+550-785 = 0$
or $r^2+9r-360 = 0$.
$r = \frac{-9\pm \sqrt{9^2+4*360}}{2*1} = \frac{-9\pm \sqrt{1521}}{2} = \frac{-9\pm39}{2} = 15,-24$
Therefore the base $r$ is $15$.
So, the answer is (A)