Let the charges on two metal spheres are $\mathrm{Q}_1$ and $\mathrm{Q}_2$ before coming in contact, then the charges are,
$$
\mathrm{Q}_1=\sigma .4 \pi \mathrm{R}^2
$$
( $\therefore \sigma=$ surface charge density same for both sphere)
$$
\mathrm{Q}_2=\sigma .4 \pi\left(2 \mathrm{R}^2\right)=4\left(\sigma .4 \pi \mathrm{R}^2\right)=4 \mathrm{Q}_1
$$
Let the charges on two metal spheres, after coming in contact becomes $\mathrm{Q}_1^{\prime}$ and $\mathrm{Q}_2^{\prime}$.
Now applying law of conservation of charges is
$$
\begin{aligned}
\mathrm{Q}_1^{\prime}+\mathrm{Q}_2^{\prime} &=\mathrm{Q}_1+\mathrm{Q}_2 \\
&=5 \mathrm{Q}_1=5\left(\sigma .4 \pi \mathrm{R}^2\right)
\end{aligned}
$$
After coming in contact, they acquire equal potentials. So, we get
$$
\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}_1^{\prime}}{\mathrm{R}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}_2^{\prime}}{\mathrm{R}}
$$
Putting the value of $Q_1^{\prime}$ and $Q_2^{\prime}$ in above equation.
$\therefore \quad Q_1^{\prime}=\frac{5}{3}\left(\sigma .4 \pi R^2\right)$
and
$\therefore \quad \sigma_1=\frac{5}{3 \sigma}$ and
$\therefore \quad \sigma_2=\frac{5}{6} \sigma$