Given,
$$
\begin{aligned}
& P(X=0)=3 C^3 \\
& P(X=2)=5 C-10 C^2 \\
& \text { and } \quad P(X=4)=4 C-1 \\
& \text { We know that, } \\
& \Sigma P(X)=1 \\
& \Rightarrow \quad 3 C^3+\left(5 C-10 C^2\right)+(4 C-1)=1 \\
& \Rightarrow \quad 3 C^3-10 C^2+9 C-2=0 \\
&
\end{aligned}
$$
$$
\begin{array}{ll}
\Rightarrow & (C-1)\left(3 C^2-7 C+2\right)=0 \\
\Rightarrow & (C-1)(3 C-1)(C-2)=0 \\
\Rightarrow & C=1, \frac{1}{3^{\prime}}, 2 \\
\therefore & C=\frac{1}{3}
\end{array}
$$

Now,


Hence, variance $=\Sigma X_P^2-\left(\Sigma X_P\right)^2$
$$
\begin{aligned}
& =\left(0^2 \times \frac{1}{9}+4 \times \frac{5}{9}+16 \times \frac{1}{3}\right)-\left(\frac{10}{9}+\frac{4}{3}\right)^2 \\
& =\left(\frac{20}{9}+\frac{16}{3}\right)-\left(\frac{66}{27}\right)^2=\frac{60+144}{27}-\frac{484}{81} \\
& =\frac{204}{27}-\frac{484}{81}=\frac{612-484}{81}=\frac{128}{81}
\end{aligned}
$$