Making use of the cube root table,

Solution:

The number $34.2$ could be written as $\frac{342}{10}$.

Now

$\sqrt[3]{34.2}=\sqrt[3]{\frac{342}{10}}=\frac{\sqrt[3]{342}}{\sqrt[3]{10}}$

Also

$340<342<350 \Rightarrow \sqrt[3]{340}<\sqrt[3]{342}<\sqrt[3]{350}$

From the cube root table, we have: 

$\sqrt[3]{340}=6.980$ and $\sqrt[3]{350}=7.047$

For the difference $(350-340)$, i.e., 10 , the difference in values

$=7.047-6.980=0.067 .$

$\therefore$ For the difference $(342-340)$, i.e., 2 , the difference in values

$=\frac{0.067}{10} \times 2=0.013$ (upto three decimal places)

$\therefore \sqrt[3]{342}=6.980+0.0134=6.993$ (upto three decimal places)

From the cube root table, we also have: 

$\sqrt[3]{10}=2.154$

$\therefore \sqrt[3]{34.2}=\frac{\sqrt[3]{342}}{\sqrt[3]{10}}=\frac{6.993}{2.154}=3.246$

Thus, the required cube root is 3.246.