Let $G(\omega) = H_1(\omega) \star H_2(\omega)$, then
$$
G(\omega) = \int_{\phi = - \pi}^{\pi} H_1(\phi) H_2(\omega - \phi) d\phi \tag{1}
$$
The nonzero ranges for $H_1$ and $H_2$ are found as:
$$
H_1(\theta) = \begin{cases} { 1 ~~~ , ~~~~ -0.2\pi < \theta < 0.2 \pi \\ 0 ~~~ , ~~~~ \text{ otherwise } } \end{cases}
$$
$$
H_2(\theta) = \begin{cases} { 1 ~~~ , ~~~~ -0.1\pi < \theta < 0.1 \pi \\ 0 ~~~ , ~~~~ \text{ otherwise } } \end{cases}
$$
Also the nonzero range of the convolution is (convolution bounds) is also known to be: $ -0.3 \pi < \omega < 0.3 \pi $.
Then the nonzero range of integration in Eq(1) for the integral is found as:
$$ \{ -\pi < \phi < \pi \} \cap \{ -0.2 \pi < \phi < 0.2 \pi \} \cap \{ -0.1 \pi < \omega - \phi < 0.1 \pi \} $$
$$ \{ -\pi < \phi < \pi \} \cap \{ -0.2 \pi < \phi < 0.2 \pi \} \cap \{ \omega -0.1 \pi < \phi < \omega + 0.1 \pi \} $$
which yields:
$$ \max(-\pi, -0.2\pi, \omega - 0.1\pi) < \phi < \min(\pi, 0.2\pi, \omega + 0.1\pi) $$
$$ \max(-0.2\pi, \omega - 0.1\pi) = \beta < \phi < \alpha =\min(0.2\pi, \omega + 0.1\pi) $$
Then the convolution integral becomes
$$
G(\omega) = \int_{\phi = \beta}^{\alpha} d\phi = \alpha - \beta \tag{2}
$$
$$
= \min(0.2\pi, \omega + 0.1\pi) - \max(-0.2\pi, \omega - 0.1\pi)
$$
Now using the range of $\omega$ as $-0.3 \pi < \omega < 0.3 \pi $ , evaluate the piece-wise result:
$$
G(\omega) = \begin{cases} {
0 ~~~~~~~~~~~~~~~~~~,~~~ -\pi < \omega < -0.3\pi \\
\omega + 0.3\pi ~~~~~,~~~ -0.3\pi < \omega < -0.1\pi \\
0.2\pi ~~~~~~~~~~~~ ,~~~ -0.1\pi < \omega < 0.1\pi \\
0.3\pi - \omega ~~~~~ ,~~~ 0.1\pi < \omega < 0.3\pi \\
0 ~~~~~~~~~~~~~~~~~~,~~~ 0.3\pi < \omega < \pi
}\end{cases}
\tag{3}
$$