Answer
${I_z} = \frac{2}{3}M{a^2}$
Work Step by Step
We have the box defined by ${\cal W} = \left[ { - a,a} \right] \times \left[ { - a,a} \right] \times \left[ {0,H} \right]$.
The volume is $V = \left( {2a} \right)\left( {2a} \right)\left( H \right) = 4{a^2}H$.
Assuming that the mass density is uniform, the mass density $\delta$ is $\delta = \frac{M}{V} = \frac{M}{{4{a^2}H}}$.
Evaluate the moment of inertia ${I_z}$:
${I_z} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \left( {{x^2} + {y^2}} \right)\delta \left( {x,y,z} \right){\rm{d}}V$
$ = \frac{M}{{4{a^2}H}}\mathop \smallint \limits_{x = - a}^a \mathop \smallint \limits_{y = - a}^a \mathop \smallint \limits_{z = 0}^H \left( {{x^2} + {y^2}} \right){\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = \frac{M}{{4{a^2}H}}\mathop \smallint \limits_{x = - a}^a \mathop \smallint \limits_{y = - a}^a \left( {{x^2} + {y^2}} \right)\left( {z|_0^H} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{M}{{4{a^2}}}\mathop \smallint \limits_{x = - a}^a \mathop \smallint \limits_{y = - a}^a \left( {{x^2} + {y^2}} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{M}{{4{a^2}}}\mathop \smallint \limits_{x = - a}^a \left( {\left( {{x^2}y + \frac{1}{3}{y^3}} \right)|_{ - a}^a} \right){\rm{d}}x$
$ = \frac{M}{{2{a^2}}}\mathop \smallint \limits_{x = - a}^a \left( {a{x^2} + \frac{1}{3}{a^3}} \right){\rm{d}}x$
$ = \frac{M}{{2a}}\left( {\left( {\frac{1}{3}{x^3} + \frac{1}{3}{a^2}x} \right)|_{ - a}^a} \right)$
$ = \frac{M}{a}\left( {\frac{{{a^3}}}{3} + \frac{{{a^3}}}{3}} \right) = \frac{2}{3}M{a^2}$
So, ${I_z} = \frac{2}{3}M{a^2}$.
