Here, 4 values are given, therefore
$$
\begin{aligned}
\Delta^{4} f(x)=0 \forall x & \\
\Rightarrow &(E-1)^{4} f(x)=0 \\
\Rightarrow &\left(E^{4}-4 E^{3}+6 E^{2}-4 E+1\right) f(x)=0 \\
\Rightarrow & E^{4} f(x)-4 E^{3} f(x)+6 E^{2} f(x) \\
-4 E f(x)+f(x)=0 \\
\Rightarrow & f(x+4)-4 f(x+3)+6 f(x+2) \\
-4 f(x+1)+f(x)=0
\end{aligned}
$$
On putting $x=0$, we get
$$
f(4)-4 f(3)+6 f(2)-4 f(1)+f(0)=0 \quad \ldots(\mathrm{i})
$$
On substituting the values of $f(0), f(1), f(2), f(4)$ in Eq. (i), we get
$$
\begin{array}{ll}
& 81-4 f(3)+6 \times 9-4 \times 3+1=0 \\
\Rightarrow & 4 f(3)=124
\end{array}
$$
$\Rightarrow \quad f(3)=31$