$\begin{aligned} & \text { } C_1: x^2+y^2+x+y-4=0 \\ & C_2: 3 x^2+3 y^2-x-y-k=0 \\ & \Rightarrow \quad x^2+y^2-\frac{x}{3}-\frac{y}{3}-\frac{k}{3}=0\end{aligned}$
Length of tangents drawn from external point to the circle is $\sqrt{S_1}$.
Now, according to the question,
Let length of tangents drawn from point $(1,2)$ to circles $C_1$ and $C_2$ are $L_1$ and $L_2$ respectively.
So,
$$
\begin{aligned}
L_1 & =\sqrt{S_1}=\sqrt{1^2+2^2+1+2-4} \\
& =\sqrt{1+4+3-4}=\sqrt{4} \\
L_2 & =\sqrt{S_1^{\prime}}=\sqrt{1^2+2^2-\frac{1}{3}-\frac{2}{3}-\frac{k}{3}} \\
& =\sqrt{1+4-1-\frac{k}{3}}=\sqrt{4-\frac{k}{3}}
\end{aligned}
$$
Now, given $\frac{L_1}{L_2}=\frac{4}{3}$
$$
\begin{array}{rlrl}
\Rightarrow & & \frac{\sqrt{4}}{\sqrt{4-\frac{k}{3}}} & =\frac{4}{3} \\
\Rightarrow & & \frac{4}{4-\frac{k}{3}} & =\frac{16}{9}[squaring] \\
\Rightarrow & & 9 & =4\left(4-\frac{k}{3}\right) \Rightarrow \frac{9}{4}=4-\frac{k}{3} \\
\frac{k}{3} & =4-\frac{9}{4} \\
\frac{k}{3} & =\frac{7}{4} \Rightarrow k=\frac{21}{4}
\end{array}
$$