Solution:
The attic seen from above has the shape of figure 1, since the meeting points of the various faces of the roof are equidistant from the edges of the base because all the faces have the same inclination with respect to the horizontal.
We observe that we can transform the attic as in figure 2 without altering its volume. It therefore suffices to calculate the volume of this new figure. We observe that the central part of the new figure is a prism with triangular base, with height $h_{pr} = 30$ meters. The base triangle is isosceles with base angles of $30^{\circ}$ and base $b_{tr}$ of length 10 meters. Therefore the height $h_{tr}$ of the triangle measures $\frac{5}{\sqrt{3}}$ meters.
The volume of the prism is therefore:

Figure 1
$$V_{pr}=\frac{1}{2} b_{tr} h_{tr} h_{pr}=\frac{750}{\sqrt{3}}\ m^{3}$$
Figure 2
The two remaining parts of the solid can be joined together to form a pyramid with a square base of side 10 meters and height $h_{\text{pir}}$ of $\frac{5}{\sqrt{3}}$ meters. The volume of the pyramid is therefore:
$$V_{\text{pir}}=\frac{1}{3} l^{2} h_{pir}=\frac{500}{3 \sqrt{3}}\ m^{3}$$
The total volume of the attic is:
$$V_{tot}=V_{pr}+V_{pir}=\frac{2750}{3 \sqrt{3}}\ m^{3}$$