As the given diagram,
$$
\begin{aligned}
&\text { Maximum voltage } \mathrm{V}_{\max }=\frac{100}{2}=50 \mathrm{~V} \\
&\text { Minimum voltage } \mathrm{V}_{\min }=\frac{20}{2}=10 \mathrm{~V}
\end{aligned}
$$
(i) Percentage of modulation, $m$
$$
\begin{aligned}
&=\left(\frac{\mathrm{V}_{\max }-\mathrm{V}_{\min }}{\mathrm{V}_{\max }+\mathrm{V}_{\min }}\right) \times 100=\left(\frac{50-10}{50+10}\right) \times 100 \\
&=\frac{40}{60} \times 100=66.67 \%
\end{aligned}
$$
(ii) Peak carrier voltage,
$$
\mathrm{V}_{\mathrm{o}}=\frac{\mathrm{V}_{\max }+\mathrm{V}_{\min }}{2}=\frac{50+10}{2}=30 \mathrm{~V}
$$
(iii) Peak value of information voltage,
$$
\begin{aligned}
\mathrm{V}_{\mathrm{om}} &=\mathrm{mV}_{\mathrm{oc}} \\
&=\frac{66.67}{100} \times 30=20 \mathrm{~V}
\end{aligned}
$$