Radius of first steel ball,
$$
r_1=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}
$$

Radius of second steel ball,
$$
r_2=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}
$$
$\therefore$ Mass of first steel ball,
$$
m_1=V_1 \times \rho=\frac{4}{3} \pi r_1^3 \rho \quad[\rho=\text { density }]
$$

Similarly, $\quad m_2=\frac{4}{3} \pi r_2^3 \rho$
$\therefore \quad \frac{m_1}{m_2}=\frac{r_1^3}{r_2^3}=\frac{\left(2 \times 10^{-2}\right)^3}{\left(4 \times 10^{-2}\right)^3}=\frac{1}{8}$
$\Rightarrow \quad m_2=8 m_1$
For elastic collision, velocity of first ball,


$$
v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right) u_1+\left(\frac{2 m_2}{m_1+m_2}\right) u_2
$$


Here, $u_1=0$ and $u_2=81 \mathrm{cms}^{-1}$
$$
\begin{aligned}
\therefore \quad v_1 & =\left(\frac{m_1-m_2}{m_1+m_2}\right) \times 0+\left(\frac{2 m_2}{m_1+m_2}\right) \times 81 \\
& =0+\frac{2 \times 8 m_1}{m_1+8 m_1} \times 81 \quad \text { [from Eq. (i)] } \\
& =\frac{16}{9} \times 81=144 \mathrm{~cm} \mathrm{~s}^{-1} \quad
\end{aligned}
$$
[from Eq. (i)]