The areas of two similar triangles are $36\text{sq.cm}$ and $81\text{sq.cm}$ respectively. If the median of the smaller triangle is $12\text{cm}$ then the corresponding median of the larger triangle is

For any two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding linear dimensions (such as sides, altitudes, or medians).
Let ${\text{Area}}_1=36\text{sq.cm}$ and its corresponding median $m_1=12\text{cm}$.
Let ${\text{Area}}_2=81\text{sq.cm}$ and its corresponding median be $m_2$.

$$\frac{{\text{Area}}_1}{{\text{Area}}_2}={\left(\frac{m_1}{m_2}\right)}^2$$

$$\frac{36}{81}={\left(\frac{12}{m_2}\right)}^2$$

Taking the square root on both sides:

$$\frac{6}{9}=\frac{12}{m_2}$$

$$\frac{2}{3}=\frac{12}{m_2}$$

$$2\times m_2=36⟹m_2=18\text{cm}$$