A room is shaped like a …

Let the radius of the hemispherical dome be r and the total height of the building be h.
Since, the internal diameter of the dome is equal to its total height
2r = h
{tex}\Rightarrow r = \frac { h } { 2 }{/tex}
Let H be the height of the cylindrical position.
{tex}\Rightarrow H = h - r = h - \frac { h } { 2 } = \frac { h } { 2 }{/tex}
Volume of  air inside the building = Volume of air inside the dome + Volume of air inside the cylinder
{tex}\Rightarrow 41 \frac { 19 } { 21 } = \frac { 2 } { 3 } \pi r ^ { 3 } + \pi r ^ { 2 } H{/tex}
{tex}\Rightarrow \frac { 880 } { 21 } = \pi r ^ { 2 } \left( \frac { 2 } { 3 } r + H \right){/tex}
{tex}\Rightarrow \frac { 880 } { 21 } = \frac { 22 } { 7 } \times \left( \frac { h } { 2 } \right) ^ { 2 } \left( \frac { 2 } { 3 } \times \frac { h } { 2 } + \frac { h } { 2 } \right){/tex}
{tex}\Rightarrow \frac { 880 \times 7 } { 22 \times 21 } = \frac { h ^ { 2 } } { 4 } \times \left( \frac { h } { 3 } + \frac { h } { 2 } \right){/tex}
{tex}\Rightarrow \frac { 40 \times 4 } { 3 } = h ^ { 2 } \times \left( \frac { 5 h } { 6 } \right){/tex}
{tex}\Rightarrow \frac { 40 \times 4 \times 6 } { 3 \times 5 } = h ^ { 3 }{/tex}
{tex}\Rightarrow h ^ { 3 } = 64{/tex}
{tex}\Rightarrow h = 4{/tex}
Thus, the height of the building is 4 m.