Making use of the cube root table,

Solution:

We have:

$37800=2^{3} \times 3^{3} \times 175 \Rightarrow \sqrt[3]{37800}=\sqrt[3]{2^{3} \times 3^{3} \times 175}=6 \times \sqrt[3]{175}$

Also

$170<175<180 \Rightarrow \sqrt[3]{170}<\sqrt[3]{175}<\sqrt[3]{180}$

From cube root table, we have: 

$\sqrt[3]{170}=5.540$ and $\sqrt[3]{180}=5.646$

For the difference $(180-170)$, i.e., 10 , the difference in values

$=5.646-5.540=0.106$

$\therefore$ For the difference of $(175-170)$, i.e., 5 , the difference in values

$=\frac{0.106}{10} \times 5=0.053$

$\therefore \sqrt[3]{175}=5.540+0.053=5.593$

Now

$37800=6 \times \sqrt[3]{175}=6 \times 5.593=33.558$

Thus, the required cube root is 33.558.