Given equation is
$$
x^5+15 x^4+94 x^3+305 x^2+507 x+353=0
$$

If all roots of the equation is increased by $k$, then transformed equation is
$$
\begin{aligned}
(x-k)^5+15(x-k)^4+ & 94(x-k)^3+305(x-k)^2 \\
+ & 507(x-k)+353=0 \ldots
\end{aligned}
$$
for eliminate 4 th degree term, the coefficient of $x^4$ is equal to zero.
coefficient of $x^4=-5 k+15$
So, $\quad-5 k+15=0 \Rightarrow k=3$
Put value in Eq. (i),
$$
\begin{aligned}
(x-3)^5+15(x-3)^4+94(x-3)^3 & +305(x-3)^2 \\
+ & 507(x-3)+353=0
\end{aligned}
$$
$\therefore$ Coefficient of
$$
\begin{aligned}
x= & { }^5 C_4(-3)^4+15\left({ }^4 C_3(-3)^3\right)+94\left({ }^3 C_2(-3)^2\right) \\
& +305\left({ }^2 C_1(-3)\right)+507 \\
& =5(81)+15(4(-27)]+94[3 \times 9]+305[(2(-3)]+507 \\
& =405-1620+2538-1830+507=0
\end{aligned}
$$
$\therefore$ Coefficient of $x=0$