Answer
$$\frac{1}{3}abc\left( {{a^2} + {b^2} + {c^2}} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^a {\int_0^b {\int_0^c {\left( {{x^2} + {y^2} + {z^2}} \right)dx} } dy} dz \cr
& \int_0^a {\int_0^b {\left[ {\int_0^c {\left( {{x^2} + {y^2} + {z^2}} \right)dx} } \right]} dy} dz \cr
& {\text{Integrate with respect to }}x \cr
& \int_0^c {\left( {{x^2} + {y^2} + {z^2}} \right)dx} = \left[ {\frac{1}{3}{x^3} + x{y^2} + x{z^2}} \right]_0^c \cr
& = \frac{1}{3}{c^3} + c{y^2} + c{z^2} \cr
& \int_0^a {\int_0^b {\left[ {\int_0^c {\left( {{x^2} + {y^2} + {z^2}} \right)dx} } \right]} dy} dz = \int_0^a {\int_0^b {\left( {\frac{1}{3}{c^3} + c{y^2} + c{z^2}} \right)} dy} dz \cr
& \int_0^a {\int_0^b {\left( {\frac{1}{3}{c^3} + c{y^2} + c{z^2}} \right)} dy} dz = \int_0^a {\left[ {\int_0^b {\left( {\frac{1}{3}{c^3} + c{y^2} + c{z^2}} \right)} dy} \right]} dz \cr
& {\text{Integrate with respect to }}y \cr
& \int_0^b {\left( {\frac{1}{3}{c^3} + c{y^2} + c{z^2}} \right)} dy = \left[ {\frac{1}{3}{c^3}y + \frac{1}{3}c{y^3} + cy{z^2}} \right]_0^b \cr
& = \frac{1}{3}b{c^3} + \frac{1}{3}{b^3}c + bc{z^2} \cr
& \int_0^a {\left[ {\int_0^b {\left( {\frac{1}{3}{c^3} + c{y^2} + c{z^2}} \right)} dy} \right]} dz = \int_0^a {\left( {\frac{1}{3}b{c^3} + \frac{1}{3}{b^3}c + bc{z^2}} \right)} dz \cr
& {\text{Integrate}} \cr
& {\text{ = }}\left[ {\frac{1}{3}b{c^3}z + \frac{1}{3}{b^3}cz + \frac{1}{3}bc{z^3}} \right]_0^a \cr
& = \frac{1}{3}ab{c^3} + \frac{1}{3}a{b^3}c + \frac{1}{3}{a^3}bc \cr
& = \frac{1}{3}abc\left( {{a^2} + {b^2} + {c^2}} \right) \cr} $$
