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Solution

$L_1$ has equal number of first $0$'s and $1$'s, so it is CFL.

$L_2$ has equal number of $1$'s and last $0$'s, so it is also CFL.

Now,

$L_1=\{0^n1^n0^m\}$

$L_2=\{0^n1^m0^m\}$

In $L_1\cap L_2$, number of first $0$'s $=$ number of $1$'s and number of $1$'s $=$ number of last $0$'s.

So,

$L_1\cap L_2=\{0^n1^n0^n\}=L_3$

Thus, statement A is true.

$L_3=\{0^n1^n0^n\mid n\ge 1\}$ is not context free language.

So, statement B is true.

Statement C is false because $L_1$ and $L_2$ are CFL.

Statement D is false because $L_1$ contains strings like $0011000$, which are not in $L_3$.