Given observation are $x_1, x_2, x_3 \ldots \ldots x_n$.
$$
\bar{x}=\frac{x_1+x_2+x_3+\ldots . .+x_n}{n}
$$
New $\bar{x}=\frac{1}{n}\left[\frac{2 x_1+5}{3}+\frac{2 x_2+5}{3}+\ldots . .+\frac{2 x_n+5}{3}\right]$
New $\overline{\mathrm{x}}=\frac{2 \overline{\mathrm{x}}_1+5}{3}$
Mean deviation of the observations
$$
\begin{aligned}
& \frac{2 \mathrm{x}_1+5}{3}, \frac{2 \mathrm{x}_2+5}{3}, \ldots \ldots \frac{2 \mathrm{x}_{\mathrm{n}}+5}{3} \\
& \text { New M.D. }=\frac{1}{\mathrm{n}} \sum_{\mathrm{x}=1}^{\mathrm{n}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right| \\
& =\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}}\left|\frac{2 \mathrm{x}_{\mathrm{i}}+5}{3}-\frac{2 \overline{\mathrm{x}}+5}{3}\right| \\
& =\frac{1}{\mathrm{n}}\left[\left|\frac{2 \mathrm{x}_1+5-2 \overline{\mathrm{x}}-5}{3}\right|+\left|\frac{2 \mathrm{x}_2+5-2 \overline{\mathrm{x}}-5}{3}\right|+\right. \\
& \left.\ldots+\left|\frac{2 \mathrm{x}_{\mathrm{n}}+5-2 \overline{\mathrm{x}}-5}{3}\right|\right] \\
& =\left[\frac{2}{3} \frac{1}{\mathrm{n}}\left|\mathrm{x}_1-\overline{\mathrm{x}}\right|+\left|\mathrm{x}_2-\overline{\mathrm{x}}\right|+\left|\mathrm{x}_3-\overline{\mathrm{x}}\right|+\ldots .+\left|\overline{\mathrm{x}}_{\mathrm{n}}-\overline{\mathrm{x}}\right|\right] \\
& =\frac{2}{3} \times 10=\frac{20}{3}
\end{aligned}
$$
So, option (c) is correct