Answer
$I_x=I_y=I_z= \dfrac{2kL^5}{3}$
Work Step by Step
Let us consider that $I_x=\iiint_{E} (y^2+z^2) \rho(x,y,z) dV=k\int_{0}^L\int_{0}^{L}\int_0^L (y^2+z^2) dx dy dz $
and $k\int_{0}^{L} (y^2x+z^2x)_0^L dy dz=k\int_{0}^{L} [y^2(L-0)+z^2(L-0)] dy dz=(kL) \int_{0}^{L} (y^2+z^2) dy dz $
or, $(kL) \int_{0}^{L} (L^3+z^2L) dz = (kL)[\dfrac{zL^3}{3}+\dfrac{z^3L}{3}]_0^L$
and $\iiint_{E} (y^2+z^2) \rho(x,y,z) dV= \dfrac{2kL^5}{3}$
By symmetry, we have $I_x=I_y=I_z= \dfrac{2kL^5}{3}$
